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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.

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    Since solving the equation seems to be difficult, why not tell us how the equation arose. Maybe there's a way around it.2012-10-22

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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.

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    @minthao_2011 : See my answer above. Setting $Y = \sin y$, you get $4Y^4 - 2\sqrt{2}Y^3 - 2Y^2 -2\sqrt{2}Y + 1 = 0$. I plotted the left hand side of the equation in Maple from $Y = -1$ to $Y = 1$ and it was always positive, but unfortunately it got quite close to $0$.2012-10-22
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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. 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 $x = \pi + 2m\pi.$

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    @GerryMyerson : see minthao_2011's solution above. f'(t) = 5t^4 - 6t^2 + 2 > 0 for all real $t$.2012-10-23