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

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### 1 Answer

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**