How to solve this trigonometric equation?

Question

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.

Answer 1

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.

Answer 2

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 $x = \pi + 2m\pi.$