Solving this trigonometric equation: $\sqrt{3}\left(\sin x+\cos x\right)=\cos 2x-2\sin x+\sqrt{3}$

Question

Help me to solve $\sqrt{3}\left(\sin x+\cos x\right)=\cos 2x-2\sin x+\sqrt{3}$

I try to change $\cos 2x = 2\cos ^2 x-1=1-2\sin^2 x=\cos^2 x-\sin^2 x$, but I can't make any common factor.

Answer 1

You can make it in this way: $$\sqrt{3}(\sin x+\cos x)=\cos 2x-2\sin x+\sqrt{3} \Longrightarrow$$ $$\Longrightarrow \sqrt{3}(\sin x+\cos x)-\sqrt{3}=\cos^2 x - \sin^2 x - 2\sin x \Longrightarrow$$ $$\Longrightarrow \sqrt{3}(\sin x+\cos x -1)=\cos^2 x - (\sin x + 1)^2 + 1 \Longrightarrow$$ $$\Longrightarrow \sqrt{3}(\sin x+\cos x -1)=(\cos x - \sin x - 1)(\cos x + \sin x + 1) + 1$$ I think you have a typo, if I was true, we made common factor already.

If not you may be replace $\sin$ by $\pm\sqrt{1-\cos^2}$ (or equivalently $\cos$ by $\pm\sqrt{1-\sin^2}$) and solve a two quartic equation. At last exclude the roots that doesn't respect to $\pm$ sign that we choose at first.

Answer 2

This looks to be a nightmare.

For conveniency, let $s=\sin(x)$ and $c=\cos(x)$. So, we have $$\sqrt{3} (s+c)=(1-2s^2)-2s+\sqrt{3} $$ So $$c=\frac{\sqrt{3}+1-s \left(2 s+\sqrt{3}-2\right)}{\sqrt{3}}$$ Square both sides and replace $c^2$ by $1-s^2$; If I am not mistaken, this would lead to the quartic equation in $s$ $$\frac{1}{4} \left(1+2 \sqrt{3}\right)+\frac{1}{2} \left(\sqrt{3}-1\right) s+\left(\frac{3}{2}-2 \sqrt{3}\right) s^2+\left(\sqrt{3}-2\right) s^3+s^4=0$$

Now, if you are concerned by the first positive root, use Taylor series to get $$\sqrt{3}\left(\sin x+\cos x\right)-(\cos 2x-2\sin x+\sqrt{3})=-1+\left(2+\sqrt{3}\right) x+\left(2-\frac{\sqrt{3}}{2}\right) x^2+O\left(x^3\right)$$ Ignoring the higher order terms, solving the quadratic would give as an approximate solution $$x=\frac{2+\sqrt{3}-\sqrt{15+2 \sqrt{3}}}{\sqrt{3}-4}\approx 0.2491$$ while the exact solution obtained using Newton method would be $0.2520$.

I suppose that there is a typo somewhere.