Question

Number of solutions of equation, $\sin 5x\cos 3x=\sin 6x\cos 2x$, in the interval $[0,\pi ]$ is

• A. 4
• B. 5
• C. 3
• D. 2

Answer 1

Answer: (B)

The given equation can be written as
$\frac{1}{2}(\sin 8x+\sin 2x)=\frac{1}{2}(\sin 8x+\sin 4x)$
or $\sin 2x-\sin 4x$
$\Rightarrow -2\sin x\cos 3x=0$
Hence $\sin x=0$ or $\cos 3x=0$
That is, $x=n\pi (n\in I)$, or $3x=k\pi +\frac{\pi }{2}(k\in I)$
Therefore, since $x\in [0,\pi ]$, the given equation is
satisfied if $x=0,\pi ,\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6}$