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