Find the number of solutions of the equation
$sin 5 x cos 3 x = sin 6 x cos 2 x, x \in [0, π]$

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Solution

$2 sin 5 x cos 3 x = sin (5 x + 3 x) + sin (5 x - 3 x) = sin 8 x + sin 2 x$
$2 sin 6 x cos 2 x = sin (6 x + 2 x) + sin (6 x - 2 x) = sin 8 x + sin 4 x$

ie, $sin 5 x cos 3 x = sin 6 x cos 2 x \Leftrightarrow sin 8 x + sin 2 x = sin 8 x + sin 4 x$
$\Rightarrow sin 4 x = sin 2 x$

But, $sin 4 x = 2 sin 2 x cos 2 x$
$∴ sin 4 x = sin 2 x \Rightarrow 2 cos 2 x = 1$ or $sin 2 x = 0$

for $cos 2 x = \frac{1}{2}$ ,
$x = \frac{π}{6}, \frac{5 π}{6}$

for $sin 2 x = 0$
$x = 0, \frac{π}{2}, π$

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

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