Question

how to solve 2sin square theta+ root 3 cos theta+1=0

Answer 1

2(1-cos^2theta)+ root3cos theta+1=0
 on finding roots of this quadratic we get  cos theta= (root3+-3root3)/8=+-root3
theta= 5pi/6 or 7pi/6. as costheta cant b greater than 1

Answer 2

2 Sin² Ф + √3 cos Ф + 1 = 0
2 (1- cos²Ф ) + √3 cos Ф + 1 =0
multiply by -1 and rearrange
2 cos² Ф -√3 cos Ф - 3 = 0

Δ = 3 + 24 = 27

cos Ф = ( √3 +-  3 √3 ) /  4            =  √3  or  -√3 / 2

cos Ф cannot be √3 as it is > 1.

So  cos Ф = - √3/2  =>    Ф = in second and third quadrants. =  5π/6  or  7π/6
or 150 deg or 210 deg

Ф = 2nπ +-  5π/6    or  2nπ +- 7π/6