<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] HINT: use the substituion
$$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2}$$
where $$\tan\left(\frac{x}{2}\right)=t$$
then you will have
$$\frac{16t(1-t^2)}{(1+t^2)^2}-\frac{2\sqrt{6}t}{1+t^2}-\sqrt{6}\frac{(1-t^2)}{1+t^2}+1=0$$
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