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I am finding the 2nd derivative critical values for graphing a trig function. So far I have it simplified to

$$-2(\sin x+2\cos 2x)=0$$

What values for x make this equal zero? And is there a reliable way to do this, besides staring at it until I figure it out...?

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HINT:

Use Double-Angle Formula, $$\cos2x=1-2\sin^2x$$ to form a Quadratic Eqaution in $\sin x$

Keep in mind that $\displaystyle-1\le\sin x\le1$ for real $x$

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  • $\begingroup$ Now if i can just remember this for tests! $\endgroup$ – stephen Mar 25 '14 at 16:32
  • $\begingroup$ @stephen, not sure if I understand your comment $\endgroup$ – lab bhattacharjee Mar 25 '14 at 16:34

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