<s>[INST] Trigonometric equation: $2(\sin^6 x+\cos^6 x)-3(\sin^4 x+\cos^4 x)+1=0$ I'm new here, but I need your help so much to solve an equation:
$$2(\sin^6 x+\cos^6 x)-3(\sin^4 x+\cos^4 x)+1=0$$
I tried a lot like making $2[(\sin^2 x)^3 + (\cos^2 x)^3$
 [/INST] As  $\sin^2x+\cos^2x=1$
Taking square, $\displaystyle1=(\sin^2x+\cos^2x)^2=\sin^4x+\cos^4x+2\sin^2x\cos^2x$
$\displaystyle\implies \sin^4x+\cos^4x=1-2(\sin^2x\cos^2x)\ \ \ \ (1)$
Taking cube, $\displaystyle1=(\sin^2x+\cos^2x)^3=\sin^6x+\cos^6x+3\sin^2x\cos^2x(\sin^2x+\cos^2x)$
$\displaystyle\implies \sin^6x+\cos^6x=1-3(\sin^2x\cos^2x)\ \ \ \ (2)$
Equate the values of $\displaystyle\sin^2x\cos^2x$ from $(1),(2)$
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