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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main · Mathematics · Trigonometric Equations

The general solutions of ^ 2 x · x= x- x+1 is

The general solutions of sin⁡2x⋅sec⁡x=tan⁡x−sin⁡x+1\sin ^{2} x \cdot \sec x=\tan x-\sin x+1 is

Options

  1. Ax=nπ+(−1)nπ4x=n \pi+(-1)^{n} \frac{\pi}{4} or x=mπ+3π4;m,n∈zx=m \pi+\frac{3 \pi}{4} ; m, n \in z
  2. Bx=nπ+(−1)nπ2x=n \pi+(-1)^{n} \frac{\pi}{2} \quad or x=mπ+3π4;m,n∈z\quad x=m \pi+\frac{3 \pi}{4} ; \quad m, n \in z
  3. Cx=nπ+(−1)nπ2x=n \pi+(-1)^{n} \frac{\pi}{2} \quad or x=mπ+5π4;m,n∈z\quad x=m \pi+\frac{5 \pi}{4} ; m, n \in z
  4. Dx=nπ+(−1)nπ4x=n \pi+(-1)^{n} \frac{\pi}{4} or x=mπ+5π4;m,n∈zx=m \pi+\frac{5 \pi}{4} ; m, n \in z

Correct answer

B. x=nπ+(−1)nπ2x=n \pi+(-1)^{n} \frac{\pi}{2} \quad or x=mπ+3π4;m,n∈z\quad x=m \pi+\frac{3 \pi}{4} ; \quad m, n \in z

Step-by-step solution

(B) sin⁡2xsec⁡x=tan⁡x−sin⁡x+1∴sin⁡x⋅sin⁡xcos⁡x=sin⁡xcos⁡x−sin⁡x+1∴sin⁡x⋅sin⁡x=sin⁡x−sin⁡xcos⁡x+cos⁡x∴sin⁡x(sin⁡x+cos⁡x)=sin⁡x+cos⁡x∴sin⁡x(sin⁡x+cos⁡x)−(sin⁡x+cos⁡x)=0∴(sin⁡x+cos⁡x)(sin⁡x−1)=0∴sin⁡x+cos⁡x=0\sin ^{2} x \sec x=\tan x-\sin x+1\therefore \quad \frac{\sin x \cdot \sin x}{\cos x}=\frac{\sin x}{\cos x}-\sin x+1\therefore \quad \sin x \cdot \sin x=\sin x-\sin x \cos x+\cos x\therefore \sin x(\sin x+\cos x)=\sin x+\cos x\therefore \sin x(\sin x+\cos x)-(\sin x+\cos x)=0\therefore(\sin x+\cos x)(\sin x-1)=0\therefore \sin x+\cos x=0 or sin⁡x=1∴tan⁡x=−1∴x=sin⁡π+3π4\sin x=1\therefore \tan x=-1\therefore x=\sin \pi+\frac{3 \pi}{4} or x=nπ+(−1)nπ2…m,n∈Zx=n \pi+(-1)^{n} \frac{\pi}{2}\ldots \mathrm{m}, \mathrm{n} \in \mathrm{Z}
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