How to solve this trigonometric equation? I want to solve the equation $$(3-\cos 4x)(\sin x - \cos x)=2.$$
I solve by putting $t = \sin x - \cos x$, but I can not find all solutions.
 A: I have just a solution. 
\begin{equation*}
(3-\cos 4x)(\sin x - \cos x) = 2.
\end{equation*}
Note that, $3-\cos 4x > 0$, therefore $\sin x - \cos x>0$.
Put $t = \sin x - \cos x = \sqrt{2}\sin\left(x - \dfrac{\pi}{4}\right)$, and then $0<t \leqslant \sqrt{2}$. We have
$$ t^2 = (\sin x - \cos x)^2  = 1 -\sin 2x.$$
Another way, 
$$3 - \cos 4x =3 - (1-2\sin^2 2x )= 2[1-(1-t^2)^2] = 2(t^4 -2t^2 + 2)$$
The given equation has the form
$$2(t^4 -2t^2 + 2)t = 2,$$
equavalent to 
$$t^5- 2t^3+2t - 1 = 0.$$ Now we prove that this equation has only root $t = 1$.
First way. The function $f(t)=t^5 -2t^3 +2t - 1$ has $$f'(t) = 5t^4-6t^2+2>0, \forall t$$
Therefore, $f$ is an increasing function on the interval $(0; \sqrt{2}]$. And $f(1) = 0$, thus  $t = 1$ is only root.
Second way, we have
$$t^5- 2t^3+2t - 1 = 0 \Leftrightarrow (t-1)(t^4 +t^3 -t^2 -t + 1)=0.$$ Note that
$$t^4 +t^3 -t^2 -t + 1 = \biggl(t^2 + \dfrac{t}2 -1\biggr)^2 + \dfrac{3t^2}4 > 0, \forall t.$$
Thirt way, we have
$$t^4 +t^3 -t^2 -t + 1=0 \Leftrightarrow \biggl(t - \dfrac{1}{t}\biggr)^2 + \biggl(t - \dfrac{1}{t}\biggr) + 1 = 0. $$
The last equation has no sulution.
With $t = 1$, we have $\sin x - \cos x = 1$, then $x = \dfrac{\pi}2 + k2\pi$ and $x = \pi + 2m\pi.$
A: Use $\sin x - \cos x = \sqrt{2}(\sin(x - \frac{\pi}{4}))$ followed by the substitution $x = y + \frac{\pi}{4}$.  You get $(3+ \cos(4y))\sin y = \sqrt{2}$.  This is satisfied if $\sin y = \frac{1}{\sqrt{2}}$.  The equation $(3+ \cos(4y))\sin y = \sqrt{2}$ can be written as a fifth-degree polynomial in $\sin y$, and you know one of the roots, so you can get a fourth-degree polynomial.
