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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- Oct 26th 2009, 04:38 PMhazecrazeabsolute max and mins
**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. - Oct 26th 2009, 04:50 PMDebsta
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.