The number of solutions of the equation 1 + sin 4 x = cos 2 3 x ,   x ∈ - 5 π 2 , 5 π 2…

The given trigonometric equation can be written as
$1-{\mathrm{c}\mathrm{o}\mathrm{s}}^{2}3x+{\mathrm{s}\mathrm{i}\mathrm{n}}^{4}x=0$

$\Rightarrow { \mathrm{s}\mathrm{i}\mathrm{n}}^{2}3x+{\mathrm{s}\mathrm{i}\mathrm{n}}^{4}x=0$

$\Rightarrow \mathrm{s}\mathrm{i}\mathrm{n}3x=0$ and $\mathrm{s}\mathrm{i}\mathrm{n}x=0$ $, x\in \left[-\frac{5\pi }{2},\frac{5\pi }{2}\right]$

In given interval $\mathrm{s}\mathrm{i}\mathrm{n}x=0$ at $x=-2\pi , -\pi ,0,\pi , 2\pi$ and these values are also satisfied by $\mathrm{s}\mathrm{i}\mathrm{n}3x=0.$
Hence total number of solutions is $5.$