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

Rearrange 
$$\dfrac{\cos 5 \theta}{\sin \theta} + \dfrac{\sin 5 \theta}{\cos \theta} = 2$$
to get
$$\cos(4\theta) = \sin(2\theta)$$
Then, use $\cos 2x=1-2\sin^2 x$
$$1-2\sin^2(2\theta)= \sin(2\theta)$$
$$\sin(2\theta) = -1,  \space \sin(2\theta)=1/2$$
$$\theta = 15^\circ,\space \theta=75^\circ$$
