I need help solving: $\cos(4x)=\sin(2x)$

Question

In need of help solving the equation; $\cos(4x)=\sin(2x)$. I have tried re-writing $\sin(2x)$, but I'm stuck on what to do with $\cos(4x)$

Answer 1

Hint:

$\cos 4x = 1- 2\sin^2 (2x)$

Substitute this and solve the quadratic in $\sin 2x$.

Alternatively, write $\sin(2x)$ as $\cos(\dfrac{\pi}{2}-2x)$ and then use the general formula for $\cos x = \cos \alpha$ i.e. $x= 2n\pi \pm \alpha$

Answer 2

use that $$\cos(x)-\sin(y)=2 \sin \left(-\frac{x}{2}-\frac{y}{2}+\frac{\pi }{4}\right) \sin \left(\frac{x}{2}-\frac{y}{2}+\frac{\pi }{4}\right)$$

Answer 3

In general: $$\cos\alpha=\cos\beta\iff\alpha=\pm\beta+2n\pi\text{ where }n\in\mathbb Z\tag1$$

Now observe that $$\cos4x=\sin2x=\cos\left(\frac12\pi-2x\right)$$ and apply $(1)$ with $\alpha=4x$ and $\beta=\frac12\pi-2x$.