Question

The general solution of $\sin 4x\cos 2x=\cos 5x\sin x$ is

• A. $\left\{\frac{n\pi }{3}\right\}\cup \left\{\right(2n+1\left)\frac{\pi }{2}\right\},n\in Z$
• B. $\left\{\frac{n\pi }{2}\right\}\cup \left\{\right(4n+1\left)\frac{\pi }{6}\right\},n\in Z$
• C. $\left\{\frac{n\pi }{3}\right\}\cup \left\{\right(2n+1\left)\frac{\pi }{6}\right\},n\in Z$
• D. $\left\{\frac{n\pi }{2}\right\}\cup \left\{\right(4n+1\left)\frac{\pi }{6}\right\},n\in Z$

Answer 1

The correct option is A $\left\{\frac{n\pi }{3}\right\}\cup \left\{\right(2n+1\left)\frac{\pi }{2}\right\},n\in Z$

$\sin 4x\cos 2x=\cos 5x\sin x\Rightarrow 2\sin 4x\cos 2x=2\cos 5x\sin x$
$\Rightarrow \sin 6x+\sin 2x=\sin 6x-\sin 4x$
$\Rightarrow \sin 2x+\sin 4x=0$
$\Rightarrow 2\sin 3x\cos x=0$
$\Rightarrow \sin 3x=0\text{or}\cos x=0$
$i.e.,3x=n\pi \text{or}x=(2n+1)\frac{\pi }{2}$
$\Rightarrow x=\frac{n\pi }{3}\text{or}x=(2n+1)\frac{\pi }{2},n\in Z$