SOLUTION: solve 2sin^2x+3cosx-3=0

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Question 436994: solve
2sin^2x+3cosx-3=0
Answer by Gogonati(855) About Me  (Show Source):
You can put this solution on YOUR website!
As you know %28sin%28x%29%29%5E2=1-%28cos%28x%29%29%5E2, substitute in our equation and have:
2%281-%28cos%28x%29%29%5E2%29%2B3%28cos%28x%29%29-3=0 => -2%28cos%28x%29%29%5E2%2B3%28cos%28x%29%29-1=0 =>
2%28cos%28x%29%29%5E2-3cos%28x%29%2B1=0, substitute cos(x)=y and write:
2y%5E2-3y%2B1=0,solving this quadratic equation have: y=1 and y=1/2.
We now solve two new equivalent equations: 1)cos(x)=1 and 2) cos(x)=1/2,
1) cos%28x%29=1 => x=2n%2Api.
2) cos%28x%29=1%2F2 => x=2n%2Api%2B%28pi%2F3%29 and x=2n%2Api%2B%285%2Api%2F3%29,where n
positive integer.