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[tex]\sin x+\sin^2x=1\implies\sin x=1-\sin^2x=\cos^2x[/tex]

[tex]\cos^{12}x+3\cos^{10}x+3\cos^8x+\cos^6x=\sin^6x+3\sin^5x+3\sin^4x+\sin^3x[/tex]
[tex]=\sin^3x(\sin^3x+3\sin^2x+3\sin x+1)[/tex]
[tex]=\sin^3x(\sin x+1)^3[/tex]

Let [tex]y=\sin x[/tex]. Then

[tex]\sin^2x+\sin x-1=0\iff y^2+y-1=0\implies y=-\varphi,y=\varphi-1[/tex]

where [tex]\varphi=\dfrac{1+\sqrt5}2\approx1.61803[/tex] is the golden ratio.

[tex]\begin{cases}y=-\varphi\\y=\varphi-1\end{cases}\implies\begin{cases}\sin x=-\varphi\\\sin x=\varphi-1\end{cases}[/tex]

Since [tex]|\sin x|\le1[/tex] for all real [tex]x[/tex], we can omit the first equation, leaving us with

[tex]\sin x=\varphi-1\approx0.61803[/tex]

[tex]\cos^{12}x+3\cos^{10}x+3\cos^8x+\cos^6x=\sin^3x(\sin x+1)^3[/tex]
[tex]=(\varphi-1)^3\varphi^3[/tex]
[tex]=1[/tex]

where the last equality follows from the fact that [tex]\varphi=1+\dfrac1\varphi[/tex].

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