Solving $2 \sin x \left( \sqrt{3} \cos x- \sin x \right)= \sqrt{2} - 1$

Question

I wanted to know if this equation could be solved any further please.

$$2 \sin x \left( \sqrt{3} \cos x- \sin x \right)= \sqrt{2} - 1$$

I have gone this far:

$$4 \sin x \sin(60^\circ-x)= \sqrt{2}- 1$$

Thank you

Answer 1

Hint:

$$2\sqrt{3} \sin x \cos x- 2\sin^2x = \sqrt3\sin 2x+\cos2x-1=\sqrt{2} - 1.$$

Then

$$3\sin^22x=3(1-\cos^22x)=(\sqrt2-\cos 2x)^2.$$

This is a quadratic equation in $\cos 2x$.

Answer 2

Hint: $2\sin x(\sqrt{3}\cos x - \sin x) = \sqrt{2} - 1\implies \sqrt{3}\sin(2x) - (1- \cos(2x)) = \sqrt{2} - 1\implies \sqrt{3}\sin(2x)+\cos(2x)=\sqrt{2}\implies \sin(2x+\frac{\pi}{6})= \sin(\frac{\pi}{4})$. At this point, can you continue to the finish line ….?

Answer 3

Hint:

Avoid squaring which immediately introduces extraneous roots

Method$\#1:$

Divide both sides by $\sin^2x$ and replace $\sin^2x$ with $$\dfrac1{1+\cot^2x}$$ in the right hand side to form a Quadratic Equation in $\cot x$

Method$\#2:$

Divide both sides by $\cos^2x$ to form a Quadratic Equation in $\tan x$

Answer 4

Use Werner Formulas

$$2\sin x\sin(60^\circ- x)=\cos(2x-60^\circ)-\cos60^\circ$$

The rest is too easy, right?