Question

Find the solutions of the following trigonometric equation in the interval [0, 2π).

Answer 1

Given the equation:

`4x-2\sqrt[]{3}\sin \sin x+2\sin \sin x+\sqrt[]{3}-4=0`

Simplifying the given equation:

`\begin{gathered} 4x-(2\sqrt[]{3}-2)\sin \sin x+\sqrt[]{3}-4=0 \\ 4x-(2\sqrt[]{3}-2)\sin \sin x=4-\sqrt[]{3} \\ 2x-(\sqrt[]{3}-1)\sin \sin x=\frac{(4-\sqrt[]{3})}{2}_{} \end{gathered}`