<s>[INST] 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.
 [/INST] 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?
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