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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main · Mathematics · Trigonometric Equations

If 3 x ≠ 2 x , then the general solution of ^ 2 x- 2 x=2- 2 x is

If 3cos⁡x≠2sin⁡x3 \cos x \neq 2 \sin x, then the general solution of sin⁡2x−cos⁡2x=2−sin⁡2x\sin ^{2} x-\cos 2 x=2-\sin 2 x is

Options

  1. Ax=nπ+π2,n∈Zx=n \pi+\frac{\pi}{2}, \quad n \in Z
  2. Bx=n(π2)+π,n∈Zx=n\left(\frac{\pi}{2}\right)+\pi, \quad n \in Z
  3. Cx=n(π2)+π3,n∈Zx=n\left(\frac{\pi}{2}\right)+\frac{\pi}{3}, \quad n \in Z
  4. Dx=(2n+1)π,n∈Zx=(2 n+1) \pi, \quad n \in Z

Correct answer

A. x=nπ+π2,n∈Zx=n \pi+\frac{\pi}{2}, \quad n \in Z

Step-by-step solution

Given sin⁡2x−cos⁡2x=2−sin⁡2x∴sin⁡2x−1+2sin⁡2x=2−sin⁡2x…[∵cos⁡2θ=1−2sin⁡2θ]3sin⁡2x=3−sin⁡2x⇒sin⁡2x=3(1−sin⁡2x)2sin⁡xcos⁡x=3cos⁡2x⇒cos⁡x(3cos⁡x−2sin⁡x)=0∴cos⁡x=0\sin ^{2} x-\cos 2 x=2-\sin 2 x\therefore \sin ^{2} x-1+2 \sin ^{2} x=2-\sin 2 x \quad \ldots\left[\because \cos 2 \theta=1-2 \sin ^{2} \theta\right]3 \sin ^{2} x \quad=3-\sin 2 x \Rightarrow \sin 2 x=3\left(1-\sin ^{2} x\right)2 \sin x \cos x=3 \cos ^{2} x \Rightarrow \cos x(3 \cos x-2 \sin x)=0\therefore \cos x=0 or 3cos⁡x=2sin⁡x3 \cos x=2 \sin x But 3cos⁡x≠2sin⁡x3 \cos x \neq 2 \sin x as per condition given ∴cos⁡x=0⇒\therefore \cos \mathrm{x}=0 \Rightarrow\mathrm{x}=\mathrm{n}π+π2,n∈Z\pi+\frac{\pi}{2}, \mathrm{n} \in \mathrm{Z}
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