Solution of $\sqrt{\cot(3x)+\sin^2(x)-\frac{1}{4}}+\sqrt{\sqrt{3}\cos x+\sin x-2}=\sin\left(\frac{3x}{2}\right)-\frac{1}{\sqrt{2}}$

Question

Find the principal solution of trigonomeric equation $$\sqrt{\cot(3x)+\sin^2(x)-\frac{1}{4}}+\sqrt{\sqrt{3}\cos x+\sin x-2}=\sin\left(\frac{3x}{2}\right)-\frac{1}{\sqrt{2}}$$

solution I try

$$\frac{1}{2}\sqrt{4\cot(3x)+1-2\cos(2x)}+\sqrt{2}\sqrt{\sin\left(x+45^\circ \right)-1}=\sin\left(\frac{3x}{2}\right)-\frac{1}{\sqrt{2}}$$

how I simplified that expression, help me

Answer 1

In the second square root, the maximum value of the trigonometric equation is 2. so the quantity value will be 2(to support the domain.)

by that you can find out the value of x.

Answer 2

Hint:

$$\sqrt3\cos x+\sin x=2\cos(x-30^\circ)\le2$$ for real $x$