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Solve the following for x:

$\sin(2x)=\cos(x)$ with $0\le x \le 2 \pi$

Not sure how to do this, all help appreciated.

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Hint: $$\sin{2x} = 2 \sin{x} \cos{x}$$

So $$\sin{2x} = \cos{x} \iff 2 \sin{x} \cos{x} = \cos{x} \iff \cos{x} (2 \sin{x} - 1) = 0$$

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so is this correct? {%p/2, 3*%pi/2, %pi/6, 5*%pi/2} – jayson Oct 20 '13 at 6:59
@jayson, $$\frac{5\pi}2>2\pi,$$ right? – lab bhattacharjee Oct 20 '13 at 7:52

Method $1:$

$$\sin2x=\cos x=\sin\left(\frac\pi2-x\right)$$

Use $\sin x=\sin A\implies x=n\pi+(-1)^nA$ where $n$ is any integer

to get $2x=n\pi+(-1)^n \left(\frac\pi2-x\right)$

If $n$ is even $=2r$(say) $\displaystyle 2x=2r\pi+\frac\pi2-x\iff x=\frac{(4r+1)\pi}6$

We need $\displaystyle0\le x\le 2\pi\implies 0\le \frac{(4r+1)\pi}6\le 2\pi\iff 0\le 4r+1\le 12\implies 0\le r\le2$

Similarly if $n$ is odd $=2r+1$(say) $\displaystyle 2x=(2r+1)\pi-\frac\pi2+x\iff x=\frac{(4r+1)\pi}2$

We need $\displaystyle0\le x\le 2\pi\implies 0\le\frac{(4r+1)\pi}2\le 2\pi\implies 0\le(4r+1)\le4\implies r=0$

Method $2:$ $$\cos x=\sin2x=\cos \left(\frac\pi2-2x\right)$$

Use $\cos x=\cos B\implies x=2m\pi\pm B$ where $m$ is any integer

Deal the $'+'$ and the $'-'$ sign one by one

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@jayson, there are $4$ solutions, right? – lab bhattacharjee Oct 20 '13 at 7:49

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