<s>[INST] $8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$, Solve for $x$ 
Solve for $x:0\leq x \leq \frac{\pi}{2}$
$$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$$

My attempt,
I changed it into $$1+4 \sin 2x-2\sqrt{3} \sin (x+\frac{\pi}{4})=0$$
 [/INST] This is a variation of Michael Rozenberg's answer,
$$\cos x+\sin x=\sqrt2\cos\left(\dfrac\pi4-x\right)$$
Set $\dfrac\pi4-x=y\iff x=?$
$2\sin x\cos x=\sin2x=\sin2\left(\dfrac\pi4-y\right)=\cos2y=2\cos^2y-1$
So, we have $$0=4(2\cos^2y-1)-2\sqrt3\cos y+1=8\cos^2y-2\sqrt3\cos y-3$$ which is a quadratic equation in $\cos y$
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