`2sin^2 (x) = 2 + cos(x)` Find all the solutions of the equation in the interval `[0,2pi)`.

Textbook Question

Chapter 5, 5.3 - Problem 28 - Precalculus (3rd Edition, Ron Larson).
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mathace | (Level 3) Assistant Educator

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Find all solutions for `2sin^2(x)=2+cos(x)`  in the interval [0, `2pi).`

Use the pythagorean identity `sin^2(x)+cos^2(x)=1.`

Solve for `sin^2(x)`    and the pythagorean identity would be `sin^2(x)=1-cos^2(x).`







Set each factor equal to zero and solve.






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