### Trigonometry

goals
If $\displaystyle 4\cos^{2}A-3=0$ and $\displaystyle 0^{\circ}\leq A \leq 90^{\circ}$, then find
$ABC$ is a triangle, $O$ is a point inside the triangle so that its distance from $A, B, C$ are $a, b, c$ respectively. $L, M, N$ are the feet of the perpendiculars from $O$ to $AB, BC, CA$ respectively. $x, y, z$ are respectively the distances of $O$ from $L, M, N$.
$\angle OAL = \alpha, \angle OBM = \beta, \angle OCN = \gamma$.
$x,y,z$ are respectively the sines and $p,q,r$ are respectively the cosines of the angles $\alpha, \beta, \gamma$ which are in A.P. with common difference $\dfrac{2\pi}{3}$
Value of $\theta (0 < \theta < 360^o )$ which satisfy the equation $\text{cosec }\theta +2 =0$ is:
Prove ${\cos ^2}x + {\cos ^2}\left( {x + \dfrac{\pi }{3}} \right) + {\cos ^2}\left( {x - \dfrac{\pi }{3}} \right) = \dfrac{3}{2}$