Find the absolute max and mins of f on the given interval. $\displaystyle f(x)=2cos(x) + sin(2x), [0, \frac{pi}{2}]$ $\displaystyle f'(x)=-2sin(x) + 2cos(2x)=0$ $\displaystyle cos(2x)=sin(x) $ I got stuck after this.
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Remember that cos (2x) = 1 - 2 (sin x)^2 So use that on the LHS, then you'll have a quadratic to solve fo sin x.
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