Solve for x :$\tan x-\sin x+\tan x-1=0$ where $x\in[0,2\pi]$

Question

The best progress I could make was to arrive at a degree 4 polynomial in $\cos x$: $-\cos^4 x-4\cos^2x+4\cos^3x-4\cos x+4=0$. But I couldn't factor it out so that I can find the value of $\cos x$ and thus $x$. Can anyone please help me to factor the polynomial or suggest an alternate method to solve the problem

Answer 1

The given quartic is not equivalent to the problem in the title. One may note that

$$2\tan x-\sin x+1=0\\2\frac{\sin x}{\cos x}-\sin x+1=0\\2\frac{\sin x}{\cos x}=1-\sin x\\4\frac{\sin^2x}{\cos^2x}=1-2\sin x+\sin^2x$$

Now use the Pythagorean identity to see that

$$4\frac{\sin^2x}{1-\sin^2x}=1-2\sin x+\sin^2x$$

Or,

$$4\sin^2x=1-2\sin x+2\sin^3x-\sin^4x$$

$$t=\sin x\implies0=t^4-2t^3+4t^2+2t-1$$

Now there are several approaches to factoring quartics. I won't go through the details, but you can read up on them here.

More or less, I recommend applying WolframAlpha, then extract desired solutions.

Answer 2

use $$\sin(x)=\frac{2t}{1+t^2}$$ $$\tan(x)=\frac{2t}{1-t^2}$$ and $$\tan\left(\frac{x}{2}\right)=t$$