Solve $\csc^2(x)=\csc^2(\frac x 2)+\csc^2 (\frac x 4)$ 
Solve  $\displaystyle\csc^2(x)=\csc^2\left(\frac x 2\right)+\csc^2\left(\frac x 4\right)$ , $\forall 
x\in[0,\pi/2] $.

My attempt:
$$\frac{1}{\sin^2\theta}=\frac{1}{\sin^2\frac \theta 2}+\frac{1}{\sin^2\frac \theta 4}$$
then $$1=\frac{\sin^2\theta}{\sin^2\frac \theta 2}+\frac{\sin^2\theta}{\sin^2\frac \theta 4}$$
we have $\theta=2\times\frac \theta 2$ and $\frac \theta 2=2\times\frac \theta4$ then
$$1= 4\cos^2 \frac \theta 2+16\cos^2 \frac \theta 2 \cos^2 \frac \theta 4 $$
$$1= \left(4+16\cos^2 \frac \theta 4\right)\cos^2 \frac \theta 2.$$
I'm stuck here. Any help
 A: Starting from your last equation
$$1=\left(4+16 \cos ^2\left(\frac{\theta }{4}\right)\right) \cos ^2\left(\frac{\theta
   }{2}\right)$$
$$\cos ^2\left(\frac{\theta }{4}\right)=x \implies \theta=4 \cos ^{-1}\left(\sqrt{x}\right)\implies 1=4 (1-2 x)^2 (4 x+1)$$ So, we need to solve the cubic equation
$$64 x^3-48 x^2+3=0$$ Since $\Delta=331776$, this equation has three real roots.
Using the trigonometric method, they are given by
$$x_k=\frac{1}{4}+\frac{1}{2} \sin \left(2k\frac{\pi  }{3}+\frac{\pi }{18}\right)\qquad \text{for}\qquad k=0,1,2$$ The problem is that the trigonometric functions of angles which are multiples of $\frac \pi{18}$ are given by infinite nested radicals (see here). So, numerically
$$x_0=-0.219846\cdots\qquad x_1=0.336824\cdots\qquad x_2=0.633022\cdots$$ The first root is negative, then discarded. Back to $\theta$, we have
$$\theta_1=4 \cos ^{-1}\left(\sqrt{\frac{1}{4}+\frac{1}{2} \sin \left(\frac{\pi
   }{18}\right)}\right)=3.80648\cdots$$
$$\theta_2=4 \cos ^{-1}\left(\sqrt{\frac{1}{4}+\frac{1}{2} \cos \left(\frac{2 \pi
   }{9}\right)}\right)=2.60302\cdots$$
