If $\cos{x}*\cos{2x}=\frac{1}{4}$, $x\in[0,90^o)$, then what is the solution of the equation?

Question

If $\cos{x}*\cos{2x}=\frac{1}{4}$, $x\in[0,90^o)$, then what is the solution of the equation?

I attempted to solve this question as follows:

$\cos{2x}=\cos^2{x}-\sin^2{x}$

$\implies \cos{x}(\cos^2{x}-\sin^2{x})=\frac{1}{4}$

$\cos^3{x}-\sin{x}(\cos{x}\sin{x})$

And I got stuck here, I did not know how to continue. I plugged in a few values for $x$ and worked out that $x=36^o$ is the solution. Could you please help me solve this question?

Answer 1

Since $x \neq 0$, multiplying both sides by $4\sin x$ and using double angle formula, $\sin 2\theta = 2\sin \theta \cos \theta \,$ twice, it is obtained $$\sin 4x = \sin x$$

whence $$4x+x=\pi \Rightarrow x=\pi/5=36^{\circ}$$

Answer 2

$\cos x\cdot \cos(2x) = \dfrac{1}{4} \implies 4\cos x\cdot (2\cos^2 x - 1) =1\implies 8\cos^3x-4\cos x-1=0$. Observe that $\cos x = -\frac{1}{2}$ is a solution. Can you finish it with factoring ?

Answer 3

$\cos x(2\cos^2x-1)=\frac 14=(-\frac 12)^2$

$\rightarrow\cos x=-\frac 12=\cos (\pi-\frac{\pi}3) $

$\rightarrow x=(2k+1)\pi-\frac{\pi}3$