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 $$L, M, N$$.

$$\angle OAL = \alpha, \angle OBM = \beta, \angle OCN = \gamma$$.

$$\angle OAL = \alpha, \angle OBM = \beta, \angle OCN = \gamma$$.

$$x,y,z$$ 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 $${\cos ^2}x + {\cos ^2}\left( {x + \dfrac{\pi }{3}} \right) + {\cos ^2}\left( {x - \dfrac{\pi }{3}} \right) = \dfrac{3}{2}$$