Solve $\sin x-\cos x-4\cos^2 x \sin x=4\sin^2 x$

Question

Solve the equation

$\sin x-\cos x-4\cos^2 x \sin x=4\sin^2 x$

My attempt:

I have rewritten the equation as: $$\sin x=\frac{4\cos ^2 x-\cos x-4}{4\cos^2 x-1}$$

I tried drawing graphs of LHS and RHS. But is there any analytical way?

Answer 1

Try using the half-angle substitution: $t = \tan(x/2)$. Then the following relations hold: \begin{align*} \cos(x) &= \frac{1 - t^2}{1 + t^2} \\ \sin(x) &= \frac{2t}{1 + t^2}. \end{align*} Then the equation becomes: $$\frac{2t}{1 + t^2} - \frac{1 - t^2}{1 + t^2} - 8 \frac{t(1 - t^2)^2}{(1 + t^2)^3} = 16\frac{t^2}{(1 + t^2)^2}.$$ Multiplying both sides by $(1 + t^2)^3$, we get $$(t^2 - 2t - 1)(1 + t^2)^2 - 8t(1 - t^2)^2 = 16t^2(1 + t^2).$$ When we expand and simplify, we get $$t^6 - 10 t^5 - 15 t^4 + 12 t^3 - 17 t^2 - 10 t - 1 = 0.$$ Wolfram Alpha gives no elementary factors or roots, with two being non-real complex conjugates. But, if $t$ is a root of the above polynomial, then $x = 2\tan^{-1}(t)$ will be one such solution.

Answer 2

Let $\omega=\cos(x)+i\sin(x)$, so that $\cos(x)=(\omega+\omega^{-1})/2$ and $\sin(x)=(\omega-\omega^{-1})/2i$. Then the equation we are trying to solve, when expressed in terms of $\omega$ and simplified is: $$\frac{i\omega^6+2\omega^5-\omega^4-4\omega^3-\omega^2+2\omega-i}{2\omega^3}=0.$$

Unfortunately, there are no obvious factors to this equation, although wolfram alpha suggests that 3 of the 6 complex roots are of length 1, which means they correspond to actual solutions to the original problem.