solve the equation on the interval 02pi 1 sintheta2cos2theta

solve the equation on the interval [0,2pi) 1 sin(theta)=2cos^2(theta)

Solution

Solution:

Here;

1 + sin = 2 cos²

=> 1 + sin = 2(1- sin²)

=> 2sin² + sin - 1 = 0

=> (2sin - 1)(sin + 1) = 0

=> sin = 1/2 and -1

=> = /6 , 5/6 , 3/2 for interval [0,2)