Question

The number of solution of the equation sin 5x 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

$\sin 5x\cos 3x=\sin 6x\cos 2x$ $x\in [0,\pi ]$

$\Rightarrow \frac{1}{2}$ (sin 8x + sin 2x) = $\frac{1}{2}$(sin 8x + sin 4x)

$\Rightarrow$ sin 2x - sin 4x = 0

$\Rightarrow$-2sin x cos 3x = 0

$\Rightarrow$ sin x = 0 or cos 3x = 0.

That is, $x=n\pi (n\in I)$, or $3x=2k\pi \pm \pi /2(k\in I)$.

Therefore, since $x\in [0,\pi ]$, the given equation is satisfied if $x=0,\pi ,\pi /6,\pi /2,5\pi /6$.

Hence, there are 5 solutions