Solve $\sin^6x+\cos^2x+3\sin^2x\cos^2x=\sin x$ Solve $$\sin^6x+\cos^2x+3\sin^2x\cos^2x=\sin x$$
My try:
The equation is equivalent to $$\left(\sin^2x\right)^3+\dfrac{1+\cos 2x}{2}+3\dfrac{1-\cos 2x}{2}\dfrac{1+\cos2x}{2}=\sin x$$ which is $$\left(\dfrac{1-\cos2x}{2}\right)^3+\dfrac{1+\cos2x}{2}+3\dfrac{1-\cos^22x}{4}=\sin x \iff\\(1-\cos2x)^3+4(1+\cos2x)+6(1-\cos^22x)=8\sin x \iff\\-\cos^32x-3\cos^22x+\cos2x+7=8\sin x$$ which makes the impression this isn't the most straight-forward approach. Any hints would be appreciated.
 A: You have\begin{multline}\sin^6x+\cos^2x+3\sin^2x\cos^2x=\sin x\iff\\\iff\sin^6x+1-\sin^2x+3\sin^2x(1-\sin^2x)-\sin x=0.\end{multline}So, let$$f(s)=s^6+1-s^2+3s^2(1-s^2)-s=s^6-3 s^4+2 s^2-s+1.$$You want to solve the equation $f(s)=0$ when $s\in[-1,1]$. It is easy to see that $f(1)=0$. So, $f(s)$ is a multiple of $s-1$; in fact, $f(s)=(s-1)\left(s^5+s^4-2 s^3-2 s^2-1\right)$. Now, let $g(s)=s^5+s^4-2 s^3-2 s^2-1$; I will show that we will always have $g(s)\leqslant-1$ in $[-1,1]$. So, let$$h(s)=g(s)+1=s^5+s^4-2 s^3-2 s^2.$$Asserting that $(\forall s\in[-1,1]):g(s)\leqslant-1$ is equivalent to asserting that$$(\forall s\in[-1,1]):h(s)\leqslant0.\label{a}\tag1$$But $h(s)=s^2(s^2-2)(s+1)$ and therefore it is clear that \eqref{a} holds. So, since $f(s)=(s-1)g(s)$, $f(s)=0\iff s=1$ (assuming that $s\in[-1,1]$). And $\sin x=1\iff x=\frac\pi2+2k\pi$, for some $k\in\Bbb Z$.
A: Use that $\cos^2x=1-\sin^2x$. Then you'll get the equation $$\sin^6x-3\sin^4x+2\sin^2x-\sin x+1=0$$ Let's $\sin x:=t, |t|\le1$. $t=1$ is an obvious solution (the only rational one), so it factors as $$(t-1)(t^5+t^4-2t^3-2t^2-1)=0$$ Look at the equation $(t^5+t^4-2t^3-2t^2-1)=0$ and go back to the substitution, $$\sin^5x+\sin^4x-2\sin^3x-2\sin^2x-1=0 \iff \\\sin^5x+\sin^4x-2\sin^3x-2\sin^2x=1$$ Note that the last equation can be written as $$\sin^2x(\sin^2x-2)(\sin x+1)=1$$ Considering $|\sin x|\le1$, what's the sign of the LHS?
