$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$, Solve for $x$

Question

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 $$1+4 \sin 2x-2\sqrt{3} \sin (x+\frac{\pi}{4})=0$$

Answer 1

Let $\sin{x}+\cos{x}=t$ and we get a quadratic equation: $$4(t^2-1)-\sqrt6t+1=0.$$

Answer 2

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 $$\frac{16t(1-t^2)}{(1+t^2)^2}-\frac{2\sqrt{6}t}{1+t^2}-\sqrt{6}\frac{(1-t^2)}{1+t^2}+1=0$$

Answer 3

Hint: Let $t=\sin x+ \cos x$ and use $1=\sin^2 x+\cos^2 x$.

Answer 4

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 $\cos y$