SOLUTION: Please solve the equation for the interval [0, 2pi]: {{{2sin^2(x)-sin(x)-1=0}}} Thank you!

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Question 983736: Please solve the equation for the interval [0, 2pi]:
2sin%5E2%28x%29-sin%28x%29-1=0

Thank you!
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Temporarily make a substitution, v=sin(x).

2v%5E2-v-1=0

%282v%2B1%29%28v-1%29=0

2sin%28x%29%2B1=0
OR
sin%28x%29-1=0

Go from those. Solution is based on one of the simple reference angles.