Solving $\frac{\cos^2\left(\frac\pi2 \cos\theta\right)}{\sin^2\theta} = 0.5$ 
$$\frac{\cos^2\left(\dfrac\pi2 \cos\theta\right)}{\sin^2\theta} = 0.5$$

I want to solve the above equation for $\theta$ in order to find its value, but I am stuck.
Could anyone enlighten me by a method to solve it?
 A: This is a transcendental equation; then no analytical solutions and numerical methods are required.
Making the problem more general, you want to solve for $x$ the equation
$$y=\cos ^2\left(\frac{\pi}{2}   \cos (x)\right) \csc ^2(x)\qquad \text{where} \qquad 0 \leq y \leq 1$$
As usual with trigonometric equations, there is an infinite number of solutions.
Suppose that you are concerned by the first root (we shall only consider positive solutions since the function is even). For an approximation, compose Taylor series around $x=0$ (I let you the intermediate steps to do). You will get
$$y=\frac{\pi ^2 }{16}x^2+\frac{\pi ^2 }{96}x^4+\left(\frac{17 \pi ^2}{11520}-\frac{\pi
   ^4}{768}\right) x^6+O\left(x^8\right)$$ Use series reversion to obtain
$$x=\frac{4 }{\pi }y^{1/2}-\frac{16 }{3 \pi ^3}y^{3/2}+\frac{32 \left(6+5 \pi
   ^2\right) }{15 \pi ^5}y^{5/2}+O\left(y^{7/2}\right)$$ Using $y=\frac 12$, this would give as an approximation
$$x \sim \frac{2 \sqrt{2} \left(4+5 \pi ^4\right)}{5 \pi ^5}\approx 0.90771$$ while the "exact" solution is $0.88944$; this is not too bad.
If you want to polish the solution, use Newton method and the iterates will be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0.9077104212 \\
 1 & 0.8894960222 \\
 2 & 0.8894396939 \\
 3 & 0.8894396932
\end{array}
\right)$$
Edit
After @Quanto's answer, the specific solution of
$$\cos(\pi x) +x^2=0$$ could have been approximated $\color{red}{1,400}$ years ago using
$$\cos(t) \simeq\frac{\pi ^2-4t^2}{\pi ^2+t^2}\qquad (-\frac \pi 2 \leq t\leq\frac \pi 2)$$ This would give
$$x^2+\frac{1-4 x^2}{x^2+1}=0$$ which is quadratic in $x^2$ leading to the beautiful
$$x \sim \frac{1}{\phi }=0.618$$ To keep this beauty, one iteration of Newton method gives
$$x=\frac{1}{\phi }+\frac{\phi ^2 \cos \left(\frac{\pi }{\phi }\right)+1}{\phi  \left(\pi  \phi  \sin
   \left(\frac{\pi }{\phi }\right)-2\right)}=0.629613$$
Another (very accurate) approximation is
$$x \sim \frac{638 \pi ^2-269 \pi-487}{434 \pi ^2 +1123 \pi +71}$$ which gives $18$ significant figures.
A: Let $x =\cos\theta$ to simplify the equation to
$$\cos(\pi x) +x^2=0$$
which has the trivial roots $\pm 1$ (excluded due to $\sin\theta \ne 0$), as well as the root that can be approximated with $\frac\pi2-\pi x+x^2 =0$, i.e.
$$x=\left(1+\sqrt{1-\frac2\pi} \right)^{-1}= 0.6239$$
(vs. the exact $ 0.6298$). Thus, the solutions are
$$\theta = 2\pi k\pm \text{arcsec} \left(1+\sqrt{1-\frac2\pi} \right) $$
