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Posted by Tina on Monday, April 12, 2010 at 11:18pm.

Solve for x in the interval [0,2pi) sin^2x+2cosx=2

sin^2x+2cosx=2 1 - cos^2x + 2cosx - 2 = 0 cos^2x - 2cosx + 1 = 0 (cosx - 1)^2 = 0 . . . cosx = 1 Can you take it from there?

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