Question

2sin^2 x +root 3cosx+1=0

Answer 1

2sin2x +cos x +1 =0

2sin2x +root3cos x +1 =0

2-2cos2x+root 3 cos x + 1=0

-2cos2x +root 3 cos x +3 = 0

 

Sum = root 3

Prdct= -6

-2 cos2x +2 root 3 cos x- root 3 cos x+ 3 =0

-2cosx(cosx - root 3) - root 3  (cos x- root 3 )=0

(cosx - root 3)( -2cosx- root 3 )=0

Cos x = root 3

(not possible)

Therefore x =cos  -1[root 3]

(cos inverse root 3 )

 

 

Cosx= (root 3)/2

Cos π/6 =( root 3)/2

Cos x =cos y

 

Therefore x=2nπ±π/6

Answer 2

Here is your answer!!