Solve 2cos^2(x) + 3sin(x) - 3 = 0 for x in the range [0 , 2pi]

We have to solve 2*(cos x)^2 + 3*sin x - 3 = 0 for x in
[0, 2*pi]

2*(cos x)^2 + 3*sin x - 3 =
0

use (cos x)^2 = 1 - (sin
x)^2

=> 2 - 2*(sin x)^2 + 3sin x - 3 =
0

=> 2*(sin x)^2 - 3*sin x + 1 =
0

=> 2*(sin x)^2 - 2*sin x - sin x + 1 =
0

=> 2*sin x( sin x - 1) -1 (sin x - 1) =
0

=>(2*sin x - 1)(sin x - 1) =
0

2*sin x - 1 = 0

=>
sin x = 1/2

=> x = arc sin (1/2) = pi/6 and
5*pi/6

sin x - 1 = 0

=>
sin x = 1

=> x = arc sin
(1)

=> x =
pi/2

The required values of x are x = pi/2, x
= pi/6 and x = 5*pi/6