24. Trigonometrical Equations and their General Solutions Answers

  1. Given equation is sinθ=1\sin\theta = -1

    sinθ=sin(π2)\Rightarrow \sin\theta = \sin\left(-\frac{\pi}{2}\right)

    θ=nπ+(1)n(π2)\Rightarrow \theta = n\pi + (-1)^n\left(-\frac{\pi}{2}\right)

    θ=nπ+(1)n+1π2\theta = n\pi + (-1)^{n +1}\frac{\pi}{2} where nI.n\in I.

  2. Given equation is cosθ=12\cos\theta = -\frac{1}{2}

    cosθ=cos2π3θ=2nπ±2π3\cos\theta = \cos\frac{2\pi}{3} \Rightarrow \theta = 2n\pi \pm \frac{2\pi}{3} where nI.n\in I.

  3. Given equation is tanθ=3\tan\theta = -\sqrt{3}

    tanθ=tan(π3)θ=nπ+(π3)\Rightarrow \tan\theta = \tan\left(-\frac{\pi}{3}\right) \Rightarrow \theta = n\pi + \left(-\frac{\pi}{3}\right)

    =nππ3= n\pi - \frac{\pi}{3} where nI.n\in I.

  4. Given equation is secθ=2\sec\theta = -\sqrt{2}

    secθ=sec3π4θ=2nπ±3π4\Rightarrow \sec\theta = \sec\frac{3\pi}{4}\Rightarrow \theta = 2n\pi\pm \frac{3\pi}{4} where xIx\in I

  5. Given equation is sin8θ=sinθsin9θsinθ=0\sin8\theta = \sin\theta \Rightarrow \sin9\theta - \sin\theta = 0

    2cos5θ.sin4θ=0\Rightarrow 2\cos5\theta.\sin4\theta = 0

    Either cos5θ=0\cos5\theta = 0 or sin4θ=0\sin4\theta = 0

    5θ=(2n+1)π2\Rightarrow 5\theta = (2n + 1)\frac{\pi}{2} or 4θ=nπ4\theta = n\pi

    θ=nπ4,(2n+1)π10\theta = \frac{n\pi}{4}, (2n + 1)\frac{\pi}{10} where nI.n\in I.

  6. Given equation is sin5x=cos2x\sin5x = \cos2x

    cos2x=cos(π25x)\Rightarrow \cos2x = \cos\left(\frac{\pi}{2} - 5x\right)

    2x=2nπ±(π25x)2x = 2n\pi \pm \left(\frac{\pi}{2} - 5x\right)

    x=(4n+1)π14,(4n1)π6x = (4n + 1)\frac{\pi}{14}, -(4n - 1)\frac{\pi}{6} where nIn\in I

  7. Given equation is sin3x=sinxsin3xsinx=0\sin3x = \sin x \Rightarrow \sin3x - \sin x= 0

    cos2x.sinx=0\Rightarrow \cos2x.\sin x = 0

    Either cos2x=0\cos2x = 0 or sinx=0\sin x= 0

    2x=(2n+1)π2\Rightarrow 2x = (2n + 1)\frac{\pi}{2} or x=nπx = n\pi

    x=nπ,(2n+1)π4x = n\pi, (2n + 1)\frac{\pi}{4} where nIn\in I

  8. Given equation is sin3x=cos2xcos2x=cos(π23x)\sin3x = \cos2x \Rightarrow \cos2x = \cos\left(\frac{\pi}{2} - 3x\right)

    2x=2nπ±(π23x)\Rightarrow 2x = 2n\pi \pm \left(\frac{\pi}{2} - 3x\right)

    x=2nπ5+π10,2nπ+π2x = \frac{2n\pi}{5} + \frac{\pi}{10}, -2n\pi + \frac{\pi}{2}

  9. Given equation is sinax+cosbx=0\sin ax + \cos bx = 0

    sinax+sin(π2bx)=0\Rightarrow \sin ax + \sin\left(\frac{\pi}{2} - bx\right) = 0

    2sin(π4+(ab)x2)cos((a+b)x2π4)=0\Rightarrow 2\sin\left(\frac{\pi}{4} + \frac{(a - b)x}{2}\right)\cos\left(\frac{(a + b)x}{2} - \frac{\pi}{4}\right) = 0

    Either sin(π4+(ab)x2)=0\Rightarrow \sin\left(\frac{\pi}{4} + \frac{(a - b)x}{2}\right) = 0 or cos((a+b)x2π4)=0\cos\left(\frac{(a + b)x}{2} - \frac{\pi}{4}\right) = 0

    π4+(ab)x2=nπ\Rightarrow \frac{\pi}{4} + \frac{(a - b)x}{2} = n\pi or (a+b)x2π4=(2n+1)π2\frac{(a + b)x}{2} - \frac{\pi}{4} = (2n + 1)\frac{\pi}{2}

    x=2nππ2ab,(2n+1)π+π2a+bx = \frac{2n\pi - \frac{\pi}{2}}{a - b}, \frac{(2n + 1)\pi + \frac{\pi}{2}}{a + b}

  10. Given tanxtan4x=1sinxsin4x=cosxcos4x\tan x\tan 4x =1 \Rightarrow \sin x\sin4x = \cos x\cos4x

    cosxcos4xsinxsin4x=0\Rightarrow \cos x\cos4x - \sin x\sin 4x = 0

    cos5x=05x=(2n+1)π2x=(2n+1)π10\cos 5x = 0 \Rightarrow 5x = (2n + 1)\frac{\pi}{2} \Rightarrow x = \frac{(2n + 1)\pi}{10}

  11. Given equation is cosθ=sin105+cos105\cos\theta = \sin105^\circ + \cos 105^\circ

    sin105=sin(60+45)=3+122\sin105^\circ = \sin(60^\circ + 45^\circ) = \frac{\sqrt{3} + 1}{2\sqrt{2}}

    cos105=cos(60+45)=1322\cos105^\circ = \cos(60^\circ + 45^\circ) = \frac{1 - \sqrt{3}}{2\sqrt{2}}

    cosθ=12θ=2nπ±π4\Rightarrow \cos\theta = \frac{1}{\sqrt{2}} \Rightarrow \theta = 2n\pi \pm \frac{\pi}{4}

  12. Given equation is 7cos2θ+3sin2θ=47\cos^2\theta + 3\sin^2\theta = 4

    4cos2θ+3=4cosθ=±12\Rightarrow 4\cos^2\theta + 3 = 4 \Rightarrow \cos\theta = \pm \frac{1}{2}

    If cosθ=12θ=2nπ±π3\cos\theta = \frac{1}{2}\Rightarrow \theta = 2n\pi\pm\frac{\pi}{3}

    If cosθ=12θ=2nπ±2π3\cos\theta = -\frac{1}{2}\Rightarrow \theta = 2n\pi\pm\frac{2\pi}{3}

  13. Given equation is 3tan(θ15)=tan(θ+15)3\tan(\theta - 15^\circ) = \tan(\theta + 15^\circ)

    tan(θ+15)tan(θ15)=31\Rightarrow \frac{\tan(\theta + 15^\circ)}{\tan(\theta - 15^\circ)} = \frac{3}{1}

    Applying componendo and dividendo

    tan(θ+15)+tan(θ15)tan(θ+15)tan(θ15)=42\Rightarrow \frac{\tan(\theta + 15^\circ) + \tan(\theta - 15^\circ)}{\tan(\theta + 15^\circ) - \tan(\theta - 15^\circ)} = \frac{4}{2}

    sin(θ+15+θ15)sin(θ+15θ+15)=2\Rightarrow \frac{\sin(\theta + 15^\circ + \theta - 15^\circ)}{\sin(\theta + 15^\circ - \theta + 15^\circ)} = 2

    sin2θ=22θ=nπ+(1)nπ2θ=nπ2+(1)nπ4\Rightarrow \sin2\theta = 2 \Rightarrow 2\theta = n\pi + (-1)^n\frac{\pi}{2} \Rightarrow \theta = \frac{n\pi}{2} + (-1)^n\frac{\pi}{4}

  14. Given equation is tanx+cotx=2tan2x2tanx+1=0\tan x + \cot x = 2 \Rightarrow \tan^2x - 2\tan x + 1 = 0

    (tanx1)2=0tanx=1x=nπ+π4\Rightarrow (\tan x - 1)^2 = 0 \Rightarrow \tan x = 1 \Rightarrow x = n\pi + \frac{\pi}{4}

  15. Given equation is sin2θ=sin2αsinθ=±sinα\sin^2\theta = \sin^2\alpha \Rightarrow \sin\theta = \pm \sin\alpha

    θ=nπ±α\theta = n\pi \pm\alpha

  16. Given equation is tan2x+cot2x=2\tan^2x + \cot^2x = 2

    tan4x2tan2x+1=0(tan2x1)2=0\Rightarrow \tan^4x - 2\tan^2x + 1 = 0 \Rightarrow (\tan^2x - 1)^2 = 0

    tanx=±x=nπ±π4\tan x = \pm \Rightarrow x = n\pi \pm \frac{\pi}{4}

  17. Given equation is tan2x=3cosec2x1\tan^2x = 3\cosec^2x - 1

    tan2x=2+3cot2xtan4x2tan2x3=0\Rightarrow \tan^2x = 2 + 3\cot^2x \Rightarrow \tan^4x -2\tan^2x - 3 = 0

    (tan2x+1)(tan2x3)=0\Rightarrow (\tan^2x + 1)(\tan^2x - 3) = 0

    If tan2x+1=0\tan^2x + 1 = 0 then xx will become imaginary.

    tanx=±3x=nπ±π3\therefore \tan x = \pm\sqrt{3} \Rightarrow x = n\pi \pm \frac{\pi}{3}

  18. Given equation is 2sin2x+sin22x=22\sin^2x + \sin^22x = 2

    2sin2x+4sin2xcos2x=2sin2x+2sin2x(1sin2x)=1\Rightarrow 2\sin^2x + 4\sin^2x\cos^2x = 2 \Rightarrow \sin^2x + 2\sin^2x(1 - \sin^2x) = 1

    (2sin2x1)(sin2x1)=0\Rightarrow (2\sin^2x - 1)(\sin^2x - 1) = 0

    sinx=±12\Rightarrow \sin x = \pm\frac{1}{\sqrt{2}} or sinx=±1\sin x = \pm 1

    x=nπ±π4,(2n+1)π2\Rightarrow x = n\pi \pm \frac{\pi}{4}, (2n + 1)\frac{\pi}{2}

  19. Given equation is 7cos2x+3sin2x=47\cos^2 x + 3\sin^2 x = 4

    4cos2x+3=4cosx=±12\Rightarrow 4\cos^2 x + 3 = 4 \Rightarrow \cos x = \pm \frac{1}{2}

    If cosx=12x=2nπ±π3\cos x = \frac{1}{2}\Rightarrow x = 2n\pi\pm\frac{\pi}{3}

    If cosx=12x=2nπ±2π3\cos x = -\frac{1}{2}\Rightarrow x = 2n\pi\pm\frac{2\pi}{3}

  20. Given equation is 2cos2x+2sinx=22\cos2x + \sqrt{2\sin x} = 2

    2sinx=2(1cos2x)=4sin2x\Rightarrow \sqrt{2\sin x} = 2(1 - \cos2x) = 4\sin^2x

    2sinx(122sin32x)=0\Rightarrow \sqrt{2\sin x}\left(1 - 2\sqrt{2}\sin^{\frac{3}{2}}x\right) = 0

    Either:math:sin x = 0 Rightarrow x = npi where nIn\in I

    or sin32x=122sinx=12\sin^{\frac{3}{2}}x = \frac{1}{2\sqrt{2}} \Rightarrow \sin x = \frac{1}{2}

    x=nπ+(1)nπ6\Rightarrow x = n\pi + (-1)^n\frac{\pi}{6}

  21. We know that tan2x2=1cosx1+cosx\tan^2\frac{x}{2} = \frac{1 - \cos x}{1 + \cos x}

    8(1cosx1+cosx)=1+secx=1+cosxcosx\therefore 8\left(\frac{1 - \cos x}{1 + \cos x}\right) = 1 + sec x = \frac{1 + \cos x}{\cos x}

    8cosx8cos2x=(1+cosx)2\Rightarrow 8\cos x - 8\cos^2x = (1 + \cos x)^2

    9cos2x6cosx+1=0(3cosx1)2=0\Rightarrow 9\cos^2x - 6\cos x + 1 = 0 \Rightarrow (3\cos x - 1)^2 = 0

    cosx=13x=2nπ±cos113\cos x = \frac{1}{3} \Rightarrow x = 2n\pi \pm \cos^{-1}\frac{1}{3} where ninI.n in I.

    Check x2(2n+1)π2\frac{x}{2}\neq (2n + 1)\frac{\pi}{2} and cosx=0\cos x \neq = 0 else equation will be meaningless.

    x(2n+1)π\Rightarrow x\neq (2n + 1)\pi and x(2n+1)π2x\neq (2n + 1)\frac{\pi}{2}

  22. Given equation is cosxcos2xcos3x=14\cos x\cos2x\cos3x = \frac{1}{4}

    (2cosxcos3x)2cos2x=12cos4xcos2x+2cos22x1=0\Rightarrow (2\cos x\cos3x)2\cos2x = 1 \Rightarrow 2\cos4x\cos2x + 2\cos^22x - 1 = 0

    cos4x[2cos2x+1]=0\Rightarrow \cos4x[2\cos2x + 1] = 0

    If cos4x=0x=(2n+1)π8\cos4x = 0 \Rightarrow x = (2n + 1)\frac{\pi}{8}

    If 2cos2x+1=02x=2nπ±2π32\cos2x + 1 = 0 \Rightarrow 2x = 2n\pi \pm \frac{2\pi}{3}

    x=nπ±π3x = n\pi \pm \frac{\pi}{3}

  23. Given equation is tanx+tan2x+tan3x=0\tan x + \tan2x + \tan3x = 0

    tanx+tan2x+tanx+tan2x1tanxtan2x=0\Rightarrow \tan x +\tan2x + \frac{\tan x + \tan 2x}{1 - \tan x\tan 2x} = 0

    (tanx+tan2x)(1+11tanxtan2x)=0\Rightarrow (\tan x + \tan 2x)\left(1 + \frac{1}{1 - \tan x\tan 2x}\right) = 0

    If tanx+tan2x=0tanx=tan2xx=nπ2xx=nπ3\tan x + \tan 2x = 0 \Rightarrow \tan x = -\tan 2x \Rightarrow x = n\pi -2x \Rightarrow x = \frac{n\pi}{3}

    If 1+11tanxtan2x=01 + \frac{1}{1 - \tan x\tan 2x} = 0 then tanxtan2x=2\tan x\tan 2x = 2

    tan2x1tan2x=1tanx=±12\frac{\tan^2x}{1 -\tan^2x} = 1 \Rightarrow \tan x = \pm\frac{1}{\sqrt{2}}

    x=nπ±tan112x = n\pi \pm\tan^{-1}\frac{1}{\sqrt{2}}

  24. Given equation is cotxtanxcosx+sinx=0\cot x - \tan x - \cos x + \sin x = 0

    cos2xsin2xcosxsinx(cosxsinx)=0\Rightarrow \frac{\cos^2x - \sin^2x}{\cos x\sin x} - (\cos x - \sin x) = 0

    (cosxsinx)(cosx+sinxcosxsinx1)=0\Rightarrow (\cos x - \sin x)\left(\frac{\cos x + \sin x}{\cos x\sin x} - 1\right) = 0

    If cosxsinx=0tanx=1x=nπ+π4\cos x - \sin x = 0 \Rightarrow \tan x = 1\Rightarrow x = n\pi + \frac{\pi}{4}

    If cosx+sinxcosxsinx1=0\frac{\cos x + \sin x}{\cos x\sin x} - 1= 0

    cosx+sinx=cosxsinx\Rightarrow \cos x + \sin x = \cos x\sin x

    Squaring, we get 1+sin2x=14sin2x1 + \sin2x = \frac{1}{4}\sin^2x

    sin2x=2±22\Rightarrow \sin2x = 2\pm 2\sqrt{2}

    However, 2+22>12 + 2\sqrt{2} > 1 which is not possible.

    sin2x=222=sinα\Rightarrow \sin 2x = 2 - 2\sqrt{2} = \sin\alpha (let)

    x=nπ2+(1)nα2x = \frac{n\pi}{2} + \frac{(-1)^n\alpha}{2}

  25. Given equation is 2sin2x5sinxcosx8cos2x=22\sin^2x - 5\sin x\cos x - 8\cos^2x = -2

    Clearly, cosx0\cos x \neq 0 else sin2x=1\sin^2x = -1 which is not possible.

    Therefore, we can divide both sides by cos2x\cos^2x which yields

    2tan2x5tanx8=2sec2x2\tan^2x - 5\tan x -8 = -2\sec^2x

    4tan2x5tanx6=0\Rightarrow 4\tan^2x - 5\tan x - 6 = 0

    (tanx2)(4tanx+3)=0\Rightarrow (\tan x - 2)(4\tan x + 3) = 0

    Thus, x=nπ+tan12,bπ+tan1(34)x = n\pi + \tan^{-1}2, b\pi + \tan^{-1}\left(\frac{-3}{4}\right)

  26. Given equation is (1tanx)(1+sin2x)=1+tanx(1 - \tan x)(1 + \sin2x) = 1 + \tan x

    (1tanx)(1+2tanx1+tan2x)=1+tanx\Rightarrow (1 - \tan x)\left(1 + \frac{2\tan x}{1 + \tan^2x}\right) = 1 + \tan x

    (1tanx)(1+tanx)2=(1+tanx)(1+tan2x)\Rightarrow (1 - \tan x)(1 + \tan x)^2 = (1 + \tan x)(1 + \tan^2x)

    (1+tanx)[(1tanx)(1+tanx)(1+tan2x)]=0\Rightarrow (1 + \tan x)[(1 - \tan x)(1 + \tan x) - (1 + \tan^2x)] = 0

    (1+tanx)(2tan2x)=0\Rightarrow (1 + \tan x)(-2\tan^2x) = 0

    If tan2x=0tanx=0x=nπ\tan^2x = 0 \Rightarrow \tan x = 0\Rightarrow x = n\pi

    If 1+tanx=0x=nππ41 + \tan x = 0 \Rightarrow x = n\pi - \frac{\pi}{4}

    where nIn \in I

  27. Given equation is 2(cosx+cos2x)+sin2x(1+2cosx)=2sinx2(\cos x + \cos2x) + \sin2x(1 + 2\cos x) = 2\sin x

    4cos3x2cosx2+2sin5x2cosx22sinx2cosx2=0\Rightarrow 4\cos\frac{3x}{2}\cos\frac{x}{2} + 2\sin\frac{5x}{2}\cos\frac{x}{2} - 2\sin\frac{x}{2}\cos\frac{x}{2} = 0

    2cosx2[2cos3x2+sin5x2sinx2]=0\Rightarrow 2\cos\frac{x}{2}\left[2\cos\frac{3x}{2} + \sin\frac{5x}{2} - \sin\frac{x}{2}\right] = 0

    2cosx2[2cos3x2+2cos3x2sinx]=0\Rightarrow 2\cos\frac{x}{2}\left[2\cos\frac{3x}{2} + 2\cos\frac{3x}{2}\sin x\right] = 0

    4cosx2cos3x2[1+sinx]=0\Rightarrow 4\cos\frac{x}{2}\cos\frac{3x}{2}[1 + \sin x] = 0

    If cosx2=0x=(2n+1)π\cos\frac{x}{2} = 0 \Rightarrow x = (2n + 1)\pi

    If cos3x2=0x=(2n+1)π3\cos\frac{3x}{2} = 0 \Rightarrow x = (2n + 1)\frac{\pi}{3}

    If 1+sinx=0x=nπ+(1)n+1π21 + \sin x = 0 \Rightarrow x = n\pi + (-1)^{n + 1}\frac{\pi}{2}

    So the values in the given range are x=π,π2,π3,π3,πx = -\pi, -\frac{\pi}{2}, -\frac{\pi}{3}, \frac{\pi}{3}, \pi

  28. Given equation is 4cos2xsinx2sin2x=3sinx4\cos^2x\sin x - 2\sin^2x = 3\sin x

    sinx[4cos2x2sinx3]=0\Rightarrow \sin x[4\cos^2x - 2\sin x - 3] = 0

    sinx[44sin2x2sinx3]=0\Rightarrow \sin x[4 - 4\sin^2x - 2\sin x - 3] = 0

    sinx[4sin2x+2sinx1]=0\Rightarrow \sin x[4\sin^2x + 2\sin x - 1] = 0

    If sinx=0x=nπ\sin x = 0 \Rightarrow x = n\pi

    If 4sin2x+2sinx1=04\sin^2x + 2\sin x - 1 = 0

    sinx=1±54\sin x = \frac{-1\pm\sqrt{5}}{4}

    If sinx=1+54sinx=sinπ10x=nπ+(1)nπ10\sin x = \frac{-1 + \sqrt{5}}{4} \Rightarrow \sin x = \sin\frac{\pi}{10} \Rightarrow x = n\pi + (-1)^n\frac{\pi}{10}

    If sinx=154sinx=sin54=sin(3π10)\sin x = \frac{-1 - \sqrt{5}}{4}\Rightarrow \sin x = -\sin54^\circ = \sin\left(\frac{-3\pi}{10}\right)

    x=nπ+(1)n+13π10\Rightarrow x = n\pi + (-1)^{n + 1}\frac{3\pi}{10}

  29. Given equation is 2+7tan2x=3.25sec2x2 + 7\tan^2x = 3.25\sec^2x

    8+28tan2x=13sec2x=13+13tan2x\Rightarrow 8 + 28\tan^2x = 13\sec^2x = 13 + 13\tan^2x

    15tan2x=5tanx=±13\Rightarrow 15\tan^2x = 5 \Rightarrow \tan x = \pm \frac{1}{\sqrt{3}}

    x=nπ±π6\Rightarrow x = n\pi \pm \frac{\pi}{6}

  30. Given equation is cos2x+cos4x=2cosx\cos 2x + \cos 4x = 2\cos x

    cos4x+cos2x2cosx=0\Rightarrow \cos4x + \cos2x - 2\cos x = 0

    2cos3xcosxcosx=0\Rightarrow 2\cos3x\cos x - \cos x = 0

    2cosx[cos3x1]=02\cos x[\cos 3x - 1] = 0

    If cosx=0x=(2n+1)π2\cos x = 0 \Rightarrow x = (2n + 1)\frac{\pi}{2}

    If cos3x1=03x=2nπx=2nπ3\cos 3x - 1 = 0\Rightarrow 3x = 2n\pi \Rightarrow x = \frac{2n\pi}{3}

  31. Given equation is 3tanx+cotx=5cosecx3\tan x + \cot x = 5\cosec x

    3sinxcosx+cosxsinx=5sinx\Rightarrow \frac{3\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{5}{\sin x}

    sinx(3sin2x+cos2x)=5sinxcosx\Rightarrow \sin x(3\sin^2x + \cos^2x) = 5\sin x\cos x

    sinx(2sin2x5cosx+1)=0\Rightarrow \sin x(2\sin^2x - 5\cos x + 1) = 0

    sinx(2cos2x+5cosx3)=0\Rightarrow \sin x(2\cos^2x + 5\cos x - 3) = 0

    sinx(2cosx+3)(2cosx1)=0\Rightarrow \sin x(2\cos x + 3)(2\cos x - 1) = 0

    sinx0\sin x\neq 0 because that will make cosecx\cosec x and cotx.\cot x \infty.

    2cosx+302\cos x + 3 \neq 0 because 1cosx1-1\leq \cos x\leq 1

    2cosx1=0cosx=12x=2nπ±π3\therefore 2\cos x - 1 = 0\Rightarrow \cos x = \frac{1}{2} \Rightarrow x = 2n\pi \pm \frac{\pi}{3}

  32. Given equation is 2sin2x=3cosx2\sin^2x = 3\cos x

    2cos2x+3cosx2=0\Rightarrow 2\cos^2x + 3\cos x - 2 = 0

    (2cosx1)(cosx+2)=0\Rightarrow (2\cos x - 1)(\cos x + 2) = 0

    cosx21cosx1\cos x \neq 2 \because -1\leq \cos x\leq 1

    2cosx1=0x=2nπ±π3  nI\therefore 2\cos x - 1 = 0\Rightarrow x = 2n\pi \pm \frac{\pi}{3}~\forall~n\in I

    0x2πx=π3,5π30\leq x \leq 2\pi \therefore x = \frac{\pi}{3}, \frac{5\pi}{3}

  33. Given equation is sin2xcosx=14\sin^2x - \cos x = \frac{1}{4}

    4sin2x4cosx=144cos2x4cosx=1\Rightarrow 4\sin^2x - 4\cos x = 1 \Rightarrow 4 - 4\cos^2x - 4\cos x = 1

    4cos2x+4cosx3=0\Rightarrow 4\cos^2x + 4\cos x -3 = 0

    (2cosx+3)(2cosx1)=0\Rightarrow (2\cos x + 3)(2\cos x - 1) = 0

    cosx21cosx1\cos x \neq 2 \because -1\leq \cos x\leq 1

    2cosx1=0x=2nπ±π3  nI\therefore 2\cos x - 1 = 0\Rightarrow x = 2n\pi \pm \frac{\pi}{3}~\forall~n\in I

    0x2πx=π3,5π30\leq x \leq 2\pi \therefore x = \frac{\pi}{3}, \frac{5\pi}{3}

  34. Given equation is 3tan2x2sinx=03\tan^2x - 2\sin x = 0

    3sin2x2sinxcos2x=0\Rightarrow 3\sin^2x - 2\sin x\cos^2x = 0

    3sin2x2sinx+2sin3x=0\Rightarrow 3\sin^2x - 2\sin x + 2\sin^3x = 0

    sinx(2sin2x+3sinx2)=0\Rightarrow \sin x(2\sin^2x + 3\sin x - 2) = 0

    sinx(sinx+2)(2sinx1)=0\Rightarrow \sin x(\sin x + 2)(2\sin x - 1) = 0

    sinx21sinx1\sin x\neq -2 \because -1\leq \sin x\leq 1

    If sinx=0x=nπ\sin x = 0 \Rightarrow x = n\pi

    If 2sinx1=0x=nπ+(1)nπ62\sin x - 1 = 0 \Rightarrow x = n\pi + (-1)^n\frac{\pi}{6}

  35. Given equation is sinx+sin5x=sin3x\sin x + \sin5x = \sin 3x

    sin5xsin3x+sinx=0\Rightarrow \sin5x - \sin3x + \sin x = 0

    2cos4xsinx+sinx=0\Rightarrow 2\cos4x\sin x + \sin x = 0

    sinx(2cos4x+1)=0\sin x(2\cos 4x + 1) = 0

    If sinx=0x=nπ  xI\sin x = 0\Rightarrow x = n\pi~\forall~x\in I

    If 2cos4x+1=04x=2nπ±2π32\cos4x + 1 = 0 \Rightarrow 4x = 2n\pi \pm\frac{2\pi}{3}

    x=nπ2±π6  xIx = \frac{n\pi}{2}\pm \frac{\pi}{6}~\forall~x\in I

    Thus, x=0,πx = 0, \pi and x=π6,π3,2π3,5π6x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{6}

  36. Given equation is sin6x=sin4xsin2x\sin6x = \sin4x - \sin2x

    sin6x+sin2xsin4x=0\Rightarrow \sin6x + \sin2x - \sin4x = 0

    2sin4xcos2xsin4x=0\Rightarrow 2\sin4x\cos2x - \sin4x = 0

    sin4x(2cos2x1)=0\Rightarrow \sin4x(2\cos2x - 1) = 0

    If sin4x=0x=nπ4\sin4x = 0 \Rightarrow x = \frac{n\pi}{4}

    If 2cos2x1=0cos2x=122x=2nπ±π32\cos2x - 1 = 0 \Rightarrow \cos2x = \frac{1}{2}\Rightarrow 2x = 2n\pi \pm \frac{\pi}{3}

    x=nπ±π6\Rightarrow x = n\pi \pm \frac{\pi}{6}

  37. Given equation is cos6x+cos4x+cos2x+1=0\cos6x + \cos 4x + \cos 2x + 1 = 0

    2cos5xcosx+2cos2x=0\Rightarrow 2\cos5x\cos x + 2\cos^2x = 0

    2cosx(cos5x+cosx)=0\Rightarrow 2\cos x(\cos5x + \cos x) = 0

    4cosxcos2xcos3x=0\Rightarrow 4\cos x\cos2x\cos3x = 0

    If cosx=0x=2nπ±π2\cos x = 0 \Rightarrow x = 2n\pi \pm\frac{\pi}{2}

    If cos2x=0x=nπ±π4\cos2x = 0 \Rightarrow x = n\pi \pm \frac{\pi}{4}

    If cos3x=0x=2nπ3±π6\cos3x = 0 \Rightarrow x = \frac{2n\pi}{3}\pm\frac{\pi}{6}

  38. Given equation is cosx+cos2x+cos3x=0\cos x + \cos 2x + \cos 3x = 0

    (cosx+cos3x)+cos2x=0\Rightarrow (\cos x + \cos3x) + \cos 2x = 0

    2cos2xcosx+cos2x=0\Rightarrow 2\cos2x\cos x + \cos 2x = 0

    cos2x(2cosx+1)=0\Rightarrow \cos2x(2\cos x + 1) = 0

    If cos2x=0x=(2n+1)π4\cos 2x = 0 \Rightarrow x = (2n + 1)\frac{\pi}{4}

    If 2cosx+1=0x=2nπ±2π32\cos x + 1 = 0 \Rightarrow x = 2n\pi\pm\frac{2\pi}{3}

  39. Given equation is cos3x+cos2x=sin3x2+sinx2\cos3x + \cos2x = \sin\frac{3x}{2} + \sin\frac{x}{2}

    2cos5x2cosx22sinxcosx2=0\Rightarrow 2\cos\frac{5x}{2}\cos\frac{x}{2} - 2\sin x\cos\frac{x}{2} = 0

    2cosx2[cos5x2sinx]=0\Rightarrow 2\cos\frac{x}{2}\left[\cos\frac{5x}{2} - \sin x\right] = 0

    If cosx2=0x2=(n+12)π\cos\frac{x}{2} = 0 \Rightarrow \frac{x}{2} = \left(n + \frac{1}{2}\right)\pi

    x=(2n+1)πx = (2n + 1)\pi

    If cos5x2=sinx=cos(π2x)\cos\frac{5x}{2} = \sin x = \cos\left(\frac{\pi}{2} - x\right)

    5x2=2nπ±(π2x)\Rightarrow \frac{5x}{2} = 2n\pi \pm \left(\frac{\pi}{2} - x\right)

    x=(4n+1)π/7,(4n1)π/3\Rightarrow x = (4n + 1)\pi/7, (4n - 1)\pi/3

    Thus, between 00 and 2π2\pi the values of xx are π7,5π7,π,9π7,13π7.\frac{\pi}{7}, \frac{5\pi}{7}, \pi, \frac{9\pi}{7}, \frac{13\pi}{7}.

  40. Given equation is tanx+tan2x+tan3x=tanx.tan2x.tan3x\tan x+ \tan2x + \tan3x = \tan x.\tan2x.\tan3x

    tanx+tan2x=tan3x(tanxtan2x1)\Rightarrow \tan x + \tan2x = \tan3x(\tan x\tan2x - 1)

    tanx+tan2x1tanxtan2x=tan3x\Rightarrow \frac{\tan x + \tan 2x}{1 - \tan x\tan2x} = -\tan3x

    tan(x+2x)=tan3x2tan3x=0\Rightarrow \tan(x + 2x) = -\tan3x \Rightarrow 2\tan 3x = 0

    3x=nπx=nπ3\Rightarrow 3x = n\pi \Rightarrow x = \frac{n\pi}{3}

  41. Given equation is tanx+tan2x+tanxtan2x=1\tan x + \tan 2x + \tan x\tan 2x = 1

    tanx+tan2x=1tanxtan2x\Rightarrow \tan x + \tan2x = 1 - \tan x\tan 2x

    tanx+tan2x1tanxtan2x=1\Rightarrow \frac{\tan x + \tan2x}{1 - \tan x\tan2x} = 1

    tan3x=tanπ4\Rightarrow \tan 3x = \tan\frac{\pi}{4}

    3x=nπ+π4x=(4n+1)π123x = n\pi + \frac{\pi}{4} \Rightarrow x = (4n + 1)\frac{\pi}{12}

  42. Given equation is sin2x+cos2x+sinx+cosx+1=0\sin2x + \cos2x + \sin x + \cos x + 1 = 0

    2sinxcosx+2cos2x1+sinx+cosx+1=0\Rightarrow 2\sin x\cos x + 2\cos^2x - 1 + \sin x + \cos x + 1 = 0

    sinx(2cosx+1)+cosx(2cosx+1)=0\Rightarrow \sin x(2\cos x + 1) + \cos x(2\cos x + 1) = 0

    (2cosx+1)(sinx+cosx)=0\Rightarrow (2\cos x + 1)(\sin x + \cos x) = 0

    If cosx=12x=2nπ±2π3\cos x = -\frac{1}{2} \Rightarrow x = 2n\pi\pm\frac{2\pi}{3}

    If sinx+cosx=0tanx=1x=nππ4\sin x + \cos x = 0 \Rightarrow \tan x = -1 \Rightarrow x = n\pi - \frac{\pi}{4}

  43. We have to prove that sinx+sin2x+sin3x=cosx+cos2x+cos3x\sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x

    (sinx+sin3x)+sin2x=(cosx+cos3x)+cos2x\Rightarrow (\sin x + \sin 3x) + \sin 2x = (\cos x + \cos3x) + \cos 2x

    2sin2xcosx+sin2x=2cos2xcosx+cos2x\Rightarrow 2\sin2x\cos x + \sin 2x = 2\cos2x\cos x + \cos 2x

    sin2x(2cosx+1)=cos2x(2cosx+1)\Rightarrow \sin2x(2\cos x + 1) = \cos2x(2\cos x + 1)

    (2cosx+1)(sin2xcos2x)=0\Rightarrow (2\cos x + 1)(\sin2x - \cos2x) = 0

    If 2cosx+1=0x=2nπ±2π32\cos x + 1 = 0 \Rightarrow x = 2n\pi \pm \frac{2\pi}{3}

    If sin2xcos2x=0tan2x=1=tanπ4x=nπ2+π8\sin 2x - \cos2x = 0 \Rightarrow \tan2x = 1 = \tan\frac{\pi}{4} \Rightarrow x = \frac{n\pi}{2}+\frac{\pi}{8}

  44. Given equation is cos6x+cos4x=sin3x+sinx\cos6x + \cos4x = \sin3x + \sin x

    2cos5xcosx=2sin2xcosx\Rightarrow 2\cos5x\cos x = 2\sin2x\cos x

    cosx(cos5xsin2x)=0\Rightarrow \cos x(\cos5x - \sin2x) = 0

    If x=0x=2nπ±π2x = 0 \Rightarrow x = 2n\pi\pm\frac{\pi}{2}

    If cos5x=sin2xcos5x=cos(π22x)\cos5x = \sin2x \Rightarrow \cos5x = \cos\left(\frac{\pi}{2} - 2x\right)

    5x=2nπ±(π22x)\Rightarrow 5x = 2n\pi\pm \left(\frac{\pi}{2} - 2x\right)

    Taking +ve sign 7x=2nπ+π2x=(4n+1)π147x = 2n\pi + \frac{\pi}{2} \Rightarrow x = (4n + 1)\frac{\pi}{14}

    Taling -ve sign 3x=2nππ2x=(4n1)π63x = 2n\pi - \frac{\pi}{2} \Rightarrow x = (4n - 1)\frac{\pi}{6}

  45. Given equation is sec4xsec2x=2\sec4x - \sec2x = 2

    cos2xcos4x=2cos2xcos4x\Rightarrow \cos2x - \cos4x = 2\cos2x\cos4x where cos2x,cos4x0\cos2x, \cos4x\neq 0

    cos2xcos4x=cos6x+cos2x\Rightarrow \cos2x - \cos4x = \cos6x + \cos 2x

    cos6x+cos4x=02cos5xcosx=0\Rightarrow \cos6x + \cos4x = 0 \Rightarrow 2\cos5x\cos x = 0

    If cos5x=05x=2nπ±π2x=2nπ5±π10\cos 5x = 0 \Rightarrow 5x = 2n\pi\pm\frac{\pi}{2}\Rightarrow x = \frac{2n\pi}{5}\pm\frac{\pi}{10}

    If cosx=0x=2nπ±π2\cos x = 0 \Rightarrow x = 2n\pi\pm\frac{\pi}{2}

  46. Given equation is cos2x=(2+1)(cosx12)\cos2x = (\sqrt{2} + 1)\left(\cos x - \frac{1}{\sqrt{2}}\right)

    (2cos2x1)=2+12(2cosx1)\Rightarrow (2\cos^2x - 1) = \frac{\sqrt{2} + 1}{\sqrt{2}}\left(\sqrt{2}\cos x - 1\right)

    (2cosx1)(2cosx+1112)=0\Rightarrow (\sqrt{2}\cos x - 1)\left(\sqrt{2}\cos x + 1 - 1 - \frac{1}{\sqrt{2}}\right) = 0

    (2cosx1)(2cosx1)=0\Rightarrow (\sqrt{2}\cos x - 1)(2\cos x - 1) = 0

    If 2cosx1=0x=2nπ±π4\sqrt{2}\cos x - 1 = 0 \Rightarrow x = 2n\pi \pm \frac{\pi}{4}

    If 2cosx1=0x=2nπ±π32\cos x - 1 = 0 \Rightarrow x = 2n\pi\pm\frac{\pi}{3}

  47. Given equation is 5cos2x+2cos2x2+1=05\cos2x + 2\cos^2\frac{x}{2} + 1 = 0

    10cos2x5+cosx+2=0[cos2x=2cos2x1]\Rightarrow 10\cos^2x - 5 + \cos x + 2 = 0[\because \cos2x = 2\cos^2x - 1]

    10cos2x+cosx3=0\Rightarrow 10\cos^2x + \cos x - 3 = 0

    (2cosx1)(5cosx+3)=0\Rightarrow (2\cos x -1)(5\cos x + 3) = 0

    If cosx=1/2x=π3[πxπ]\cos x = 1/2 \Rightarrow x = \frac{\pi}{3} [\because -\pi \leq x\leq \pi]

    If 5cosx+3=0x=πcos1355\cos x + 3 = 0 \Rightarrow x = \pi - \cos^{-1}\frac{3}{5}

  48. Given equation is cotxtanx=secx\cot x - \tan x = sec x

    cosx(cos2xsin2x)=sinxcosx\Rightarrow \cos x(\cos^2x - \sin^2x) = \sin x\cos x

    cosx(2sin2x+sinx1)=0\Rightarrow \cos x(2\sin^2x + \sin x - 1) = 0

    cosx(2sinx1)(sinx+1)=0\Rightarrow \cos x(2\sin x - 1)(\sin x + 1) = 0

    cosx0\cos x\neq 0 and sin1\sin \neq -1 because that will render original equation meaningless.

    2sinx1=0x=nπ+(1)nπ6\therefore 2\sin x - 1= 0 \Rightarrow x = n\pi + (-1)^n\frac{\pi}{6}

  49. Given equation is 1+secx=cot2x21 + \sec x = \cot^2\frac{x}{2}

    1+cosxcosx=cos2x2sin2x2\Rightarrow \frac{1 + \cos x}{\cos x} = \frac{\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}}

    2sin2x2cos2x2=cosxcos2x2\Rightarrow 2\sin^2\frac{x}{2}\cos^2\frac{x}{2} = \cos x\cos^2\frac{x}{2}

    cos2x2(2sin2x2cosx)=0\Rightarrow \cos^2\frac{x}{2}\left(2\sin^2\frac{x}{2} - \cos x\right) = 0

    cos2x2(12cosx)=0\Rightarrow \cos^2\frac{x}{2}\left(1 - 2\cos x\right) = 0

    If cosx2=0x2=nπ+pi2x=(2n+1)π\cos\frac{x}{2} = 0 \Rightarrow \frac{x}{2} = n\pi + \frac{pi}{2} \Rightarrow x = (2n + 1)\pi

    If 12cosx=0x=2nπ±π31 - 2\cos x = 0\Rightarrow x = 2n\pi \pm \frac{\pi}{3}

  50. Given equation is cos3xcos3x+sin3xsin3x=0\cos3x\cos^3x + \sin3x\sin^3x = 0

    (4cos3x3cosx)cos3x+(3sinx4sin3x)sin3x=0\Rightarrow (4\cos^3x - 3\cos x)\cos^3x + (3\sin x - 4\sin^3x)\sin^3x = 0

    3(sin4xcos4x)4(sin6xcos6x)=0\Rightarrow 3(\sin^4x - \cos^4x) - 4(\sin^6x - \cos^6x) = 0

    3(sin2xcos2x)4(sin2xcos2x)(sin4x+cos4x+sin2xcos2x)=0\Rightarrow 3(\sin^2x - \cos^2x) - 4(\sin^2x - \cos^2x)(\sin^4x + \cos^4x + \sin^2x\cos^2x) = 0

    cos2x[3+4{sin2x(sin2x+cos2x)+cos4x}]=0\Rightarrow \cos2x[-3 + 4\{\sin^2x(\sin^2x + \cos^2x) + \cos^4x\}] = 0

    cos2x[4cos4x4cos2x+1]=0\Rightarrow \cos 2x[4\cos^4x - 4\cos^2x + 1] = 0

    cos2x(2cos2x1)2=0\Rightarrow \cos2x(2\cos^2x - 1)^2 = 0

    cos32x=0\Rightarrow \cos^32x = 0

    cos2x=0\Rightarrow \cos 2x = 0

    2x=nπ+π2x=(2n+1)π42x = n\pi + \frac{\pi}{2} \Rightarrow x = (2n + 1)\frac{\pi}{4}

  51. Given equation is sin3x+sinxcosx+cos3x=1\sin^3x + \sin x\cos x + \cos^3x = 1

    sin3x+cos3x+sinxcosx1=0\Rightarrow \sin^3x + \cos^3x + \sin x\cos x - 1 = 0

    (sinx+cosx)(sin2xsinxcosx+cos2x)+(sinxcosx1)=0\Rightarrow (\sin x + \cos x)(\sin^2x - \sin x\cos x + \cos^2x) + (\sin x\cos x - 1) = 0

    (1sinxcosx)(sinx+cosx1)=0\Rightarrow (1 - \sin x\cos x)(\sin x + \cos x - 1) = 0

    If 1sinxcosx=0sin2x=21 - \sin x\cos x = 0 \Rightarrow \sin 2x = 2 which is not possible.

    sinx+cosx1=012sinx+12cosx=12\therefore \sin x + \cos x - 1 = 0 \Rightarrow \frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x = \frac{1}{\sqrt{2}}

    cos(xπ4)=cosπ4\Rightarrow \cos\left(x - \frac{\pi}{4}\right) = \cos\frac{\pi}{4}

    xπ/4=2nπ±π/4=2nπ,2nπ+π/2\Rightarrow x - \pi/4 = 2n\pi \pm \pi/4 = 2n\pi, 2n\pi + \pi/2

  52. Given equation is sin7x+sin4x+sinx=0\sin 7x + \sin4x + \sin x = 0

    2sin4xcos3x+sin4x=0\Rightarrow 2\sin4x\cos3x + \sin4x = 0

    sin4x(2cos3x+1)=0\Rightarrow \sin4x(2\cos3x + 1) = 0

    If sin4x=0x=nπ/4x=π/4 0xπ/2\sin4x = 0 \Rightarrow x = n\pi/4 \Rightarrow x = \pi/4~\forall 0\leq x\leq\pi/2

    If cos3x=1/2x=2π9,4π9 0xπ/2\cos3x = -1/2 \Rightarrow x = \frac{2\pi}{9}, \frac{4\pi}{9}~\forall 0\leq x\leq\pi/2

  53. Given equation is sinx+3cosx=2\sin x + \sqrt{3}\cos x = \sqrt{2}

    Dividing both sides by 22 [we arrive at this no. by squaring and adding coefficients of sinx\sin x and cosx\cos x and then taking square root]

    12sinx+32cosx=12\Rightarrow \frac{1}{2}\sin x + \frac{\sqrt{3}}{2}\cos x = \frac{1}{\sqrt{2}}

    sinπ6sinx+cosπ6cosx=cosπ4\Rightarrow \sin\frac{\pi}{6}\sin x + \cos\frac{\pi}{6}\cos x = \cos\frac{\pi}{4}

    cos(xπ6)=cosπ4\Rightarrow \cos\left(x - \frac{\pi}{6}\right) = \cos\frac{\pi}{4}

    xπ6=2nπ±π4\Rightarrow x - \frac{\pi}{6} = 2n\pi\pm\frac{\pi}{4}

    x=2nπ±5π12,2nππ12\Rightarrow x = 2n\pi\pm\frac{5\pi}{12}, 2n\pi-\frac{\pi}{12}

  54. We have to find minimum value of 27cos2x.81sin2x27^{\cos2x}.81^{\sin2x}

    27cos2x.81sin2x=33cos2x+4sin2x27^{\cos2x}.81^{\sin2x} = 3^{3\cos2x + 4\sin2x}

    It will be minimum when 3cos2x+4sin2x3\cos2x + 4\sin2x