Find general solution of 2 sin square x + root 3 cos x + 1 is equals to zero
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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
I hope it helps u dear ☺️....
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
I hope it helps u dear ☺️....
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