Precalculus

Solve for x given the interval of [0,2pi).

cos^2x=2+2sinx

I got (3pi)/2 or 270 degrees..

By the way, does the notation [0,2pi) mean
0<=x<2 pi or 360?

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  1. The reason they usually use [0,2pi) is that 0 and 2pi are really the same angle. So, your solution set need not include them both.

    cos^2x-2sinx-2 = 0
    1-sin^2x-2sinx-2 = 0
    sin^2x+2sinx+1 = 0
    sinx = -1
    so, yes x = 3pi/2

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    posted by Steve

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