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So The original function was

$$f(x) = 2\sin t-\cos 2t$$ on interval $[0,2\pi]$.

Now I took the derivative which was

$$f'(x) = 2(\cos t+\sin2t)$$

Now I set it equal to zero but I am stuck not knowing how to proceed to find the critical points:

$$\cos t + \sin 2t = 0$$ (since $2$ cannot equal $0$)

So what do I do now?

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Use $\sin 2t=2\sin t\cos t$.${}{}{}{}{}$

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The double-angle formula for $\sin 2t$ tells us $\sin 2t = 2 \sin t \cos t$ (which can also be derived from the angle sum formula for $\sin$: $\sin 2t = \sin(t + t) = \cdots$.

That gives us

$$\cos t + \sin 2t = \cos t + 2\sin t \cos t = \cos t(1 + 2 \sin t) = 0$$ $$\implies \cos t = 0, \; \text{ or }\; \sin t = -\frac 12$$

Now your only task is to determine which values of $t$ satisfy $\cos t = 0$, and which values of $t$ satisfy $\sin t = -\frac 12$.

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You have been getting quite a trig workout as late :-) +1 – Amzoti Nov 13 '13 at 1:12

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