# Solve for x when 0<=x<=2pi cos2x + sinx = 0

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Solve `cos2x+sinx=0` on `0<=x<=2pi` :

Use an identity for `cos2x` ; `cos2x=1-2sin^2x` . We choose this one since the other term is `sinx` , thus all terms involve only `sinx` .

Then:

`1-2sin^2x+sinx=0` Substitute for cos2x

`2sin^2x-sinx-1=0` Rewrite in standard form.

Recognize this as a quadratic in sinx:

`(2sinx+1)(sinx-1)=0` Factor

`sinx=-1/2` or `sinx=1` Zero product property

On the given interval, `sinx=-1/2=>x=(7pi)/6,(11pi)/6` and `sinx=1=>x=(pi)/2`

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Thus the solutions are `x=(7pi)/6,(11pi)/6,"or"(pi)/2`

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** If you don't recognize the quadratic form, try a substitution. Let y=sinx. Then you have `y^2-y-1=0` which you can solve, then substitute sinx for y in the solution.

The graph of `y=cos2x+sinx` (Look for the zeros):