Q. If $3\cos{x \neq 2\sin x}$, then the general solution of $\sin^{2}{x - \cos{2x = 2 - \sin{2x}}}$ is
A) $n\pi + ( - 1)^{n}\frac{\pi}{2},\ n \in Z$
B) $\frac{n\pi}{2},\ n \in Z$
C) $(4n \pm 1)\frac{\pi}{2},\ n \in Z$
D) $(2n - 1)\pi,\ n \in Z$
Explanation
(c)
$\sin^{2}{x - \cos{2x = 2 - \sin{2x}}}$
$\Rightarrow \ \ 1 - \cos^{2}x - \left( 2\cos^{2}{x - 1} \right) = 2 - 2\sin{x\cos x}$
$\Rightarrow \ - 3\cos^{2}{x + 2\sin{x\cos{x = 0}}}$
$\Rightarrow \ \cos{x\left( 2\sin{x - 3\cos x} \right) = 0}$
$\Rightarrow \ \cos{x = 0,\ \ \ \ \ (\because\ \ 2si9n\ x - 3\cos{x \neq 0})}$
$\Rightarrow \ \ x = 2n\pi \pm \frac{\pi}{2}$
$\Rightarrow \ \ x = (4n \pm 1)\frac{\pi}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.