Express $\cos 4x$ in terms of only $\sin 2x$ and hence solve the trigonometric equation

Question

The question is to express $\cos 4x$ in terms of only $\sin 2x$ and hence solve the trigonometric equation with the restriction of $\theta \in (0^{\circ}, 135^{\circ})$

$$\frac{\cos 5 \theta}{\sin \theta} + \frac{\sin 5 \theta}{\cos \theta} = 2.$$

I have come up to this point:

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

Now I only need to express $\cos 2x$ in terms of $\sin 2x$, but for all quadrants. Regarding the trigonometric equation, I have done these steps:

\begin{align} \frac{\cos(5x)\cos(x) + \sin(5x)\sin(x)}{\sin(x)\cos(x)}=2 &\implies \cos(5x-x)=2\sin(x)\cos(x) \\ &\implies \cos(4x)=\sin(2x). \end{align}

So I believe that when I express $\cos(4x)$ in terms of $\sin 2x$ I continue with solving the equation.

Answer 1

Rearrange

$$\dfrac{\cos 5 \theta}{\sin \theta} + \dfrac{\sin 5 \theta}{\cos \theta} = 2$$

to get

$$\cos(4\theta) = \sin(2\theta)$$

Then, use $\cos 2x=1-2\sin^2 x$

$$1-2\sin^2(2\theta)= \sin(2\theta)$$

$$\sin(2\theta) = -1, \space \sin(2\theta)=1/2$$

$$\theta = 15^\circ,\space \theta=75^\circ$$

Answer 2

Multiplying by $$\sin(x)\cos(x)$$ we get $$\cos(5x)\cos(x)+\sin(5x)\sin(x)=2\sin(x)\cos(x)$$ and this is $$\cos(5x-x)=2\sin(x)\cos(x)$$ $$\cos(4x)=\sin(2x)$$ $$\cos^2(2x)-\sin^2(2x)=\sin(2x)$$

Answer 3

Hint $$\cos 4x = \cos^2 2x - \sin^2 2x = 1-\sin^2 2x - \sin^2 2x = 1-2\sin^2 2x$$

Answer 4

Since Cos4X = Sin 2X ,$$ Cos 4X = Cos (90 - 2X) $$ 4X = 2nπ $\pm$(90-2X) $$ Hence \,X = 15 , 75 $$