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

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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).`

`2sin^2(x)=2+cos(x)`

`2(1-cos^2(x))=2+cos(x)`

`2-2cos^2(x)=2+cos(x)`

`-2cos^2(x)-cos(x)=0`

`2cos^2(x)+cos(x)=0`

`cos(x)[2cos(x)+1]=0`

Set each factor equal to zero and solve.

`cos(x)=0`

`x=pi/2,x=(3pi)/2`

`2cos(x)+1=0`

`cos(x)=-1/2`

`x=(2pi)/3,x=(4pi)/3`

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