Question

The number of solutions of equation, sin5x cos 3x = sin 6x cos 2x, in the

interval [0, $\pi$] are

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

The correct option is C

5

The given equation can be written as

$\frac{1}{2}$(sin8x + sin2x) = $\frac{1}{2}$ (sin8x + sin4x)

or, sin 2x - sin 4x=0 $\Rightarrow$ - 2 sin x cos 3x = 0

Hence sin x = 0 or cos 3x = 0.

That is, x = n$\pi$, (n $\in$ Z), or 3x = (2m+1)$\frac{\pi }{2},m\in Z$ Therefore, since x $\in$ [0, $\pi$],

the given equation is satisfied if x = 0, $\pi$, $\frac{\pi }{6}$, $\frac{\pi }{2}$, $\frac{5\pi }{6}$.