Siny day.

Using the identities

$\sin(2x) = 2\sin(x)\cos(x)$ and

$\sin(3x) = 3\sin(x) - 4\sin^{3}(x)$ , the equation becomes

$\sin(x) + 4\sin(x)\cos(x) - 3\sin(x) + 4\sin^{3}(x) = 3$

$\Longrightarrow 4\sin(x)\cos(x) + 4\sin^{3}(x) - 2\sin(x) = 3$ .

Now let $f(x) = 4\sin(x)\cos(x) + 4\sin^{3}(x) - 2\sin(x)$ . We can rewrite $f(x)$ as

$2\sin(x)(2\cos(x) + 2\sin^{2}(x) - 1) = 2\sin(x)(-2\cos^{2}(x) + 2\cos(x) + 1) = 2\sin(x)(-2(\cos(x) - \frac{1}{2})^{2} + \frac{3}{2})$ ,

which can only equal $3$ when $\sin(x) = 1$ and $\cos(x) = \frac{1}{2}$ . Since these conditions can't coexist, we can conclude that $f(x) \lt 3$ for all $x$ on the given interval, and hence there are no solutions to the original equation.

Since the max value of sin x is 1, therefore the individual sin expressions must be equal to 1 to give 3 on RHS.

But then we see that sin 3x doesn't equal -1 (because it has to add up) in the given domain. So that leaves the possibility of 2 sin 2x being equal to 2 and sin x being equal to 1, for which there is no common solution.

Hence, no. of solutions = 0.