The number of solutions of equation, 5xcos3x = 6xcos2x ,~in the...

Given,

$\sin 5xcos3x = \sin 6xcos2x$

The given equation can be written as

$2\sin 5xcos3x = 2\sin 6xcos2x$

⇒ \sin 8x + \sin 2x = \sin 8x + \sin 4x; \left[∵ 2sinAcosB = \sin \left(A + B\right) + \sin \left(A - B\right)\right]$</p><p>~$⇒ \sin 2x - \sin 4x = 0$</p><p>$⇒ - 2\sin x \cos 3x = 0,$</p><p>Hence,~$\sin x = 0$~or$\cos 3x = 0 .$</p><p>$x = n\pi \left(n \in I\right)$~or~$3x = k$\pi +$\dfrac{π}{2} \left(k \in I\right)

Therefore, since $x \in \left[0, \pi\right],$the given equation is satisfied,~$x = 0, \pi, \dfrac{π}{6}, \dfrac{π}{2}$~or~$\dfrac{5\pi}{6}.$Hence number of solutions are~$5.</p></blockquote>$