What are the real values of $k$ that satisfy the trigonometric inequality?

[anemone]

Find all real $k$ such that $0<k<\pi$ and $\dfrac{8}{3\sin k-\sin 3k}+3\sin^2 k\le 5$.

 

[Saitama]

Find all real $k$ such that $0<k<\pi$ and $\dfrac{8}{3\sin k-\sin 3k}+3\sin^2 k\le 5$.

Spoiler

Since

$$\sin 3x=3\sin x-4\sin^3 x$$

the given inequality can be written as:

$$\frac{8}{4\sin^3k}+3\sin^2k \le 5 \Rightarrow \frac{2}{\sin^3k}+3\sin^2k \le 5$$

From AM-GM inequality:

$$\frac{\frac{1}{\sin^3k}+\frac{1}{\sin^3k}+\sin^2k+\sin^2k+\sin^2k}{5} \ge (1)^{1/5}$$
$$\Rightarrow \frac{2}{\sin^3k}+3\sin^2k \ge 5$$

So we only need to check the following:

$$\frac{2}{\sin^3k}+3\sin^2k=5$$

Clearly, $k=\pi/2$ is the solution.

$\blacksquare$