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I need to find the intersection of these two functions $y = \cos x$ and $y=\sin 2x$ in the interval between $0$ and $\pi$, specifically in the point highlighted in red in the picture below:

the joint graph of cosx and sin2x

I tried equating the functions: $$ \cos x = \sin 2x \\ \cos x = 2\sin x\cos x \\ 2\sin x = \frac{\cos x}{\cos x} \\ \sin x = \frac{1}{2} \\ \cos x = 1 $$ The answer has to be in radians but when I convert the results to radians it doesn't make sense visually at all: $$ \sin x = \frac{\pi}{6} \\ \cos x = 0 $$

I wonder if my strategy is correct or the way I solved the equation is not correct.

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  • $\begingroup$ You had not to simolify by $\cos(x)$. $\endgroup$ Nov 10, 2016 at 7:36

3 Answers 3

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$$ \cos(x)=\sin(2x)$$

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

$$ \Leftrightarrow \cos(x)(2\sin(x)-1)=0$$

$$ \Leftrightarrow \cos(x)=0 ~~~\sin(x)=\frac{1}{2}$$

$$ \Leftrightarrow x=\pm\frac{\pi}{2}+2\pi k ~~~ x=\pi-\frac{\pi}{6}+2k\pi $$

where $k \in Z$

Picture of intersections: enter image description here

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  • $\begingroup$ thank you! But in your answer you specify the value of $x$ for $sin(x)$. However I need to find $sin(2x)$. So in this case shouldn't $x_1 = \frac{\pi}{12} \pm {\pi}k$ and $x_2 = \frac{5\pi}{12} \pm {\pi}k$? I mean $x_1$ and $x_2$ for $y=sin(2x)$ $\endgroup$
    – Yos
    Nov 10, 2016 at 8:06
  • $\begingroup$ My bad I didn't see your domain restriction. $\endgroup$ Nov 10, 2016 at 8:09
  • $\begingroup$ ok, so is my calculation of $x_1$ and $x_2$ correct? Also the graph picture you shared is really nice. Is it a free resource? $\endgroup$
    – Yos
    Nov 10, 2016 at 8:17
  • $\begingroup$ yes search up Desmos on google! @Yos $\endgroup$ Nov 10, 2016 at 21:26
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We have

$\cos(x)=\sin(2x) \iff$

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

$\cos(x)=0\;$ or $\;\sin(x)=\frac{1}{2} \iff$

$$x=\frac{\pi}{2}+k\pi$$ or

$$x=\frac{\pi}{6}+2k\pi$$ or

$$x=\pi-\frac{\pi}{6}+2k\pi.$$

So, in $[0,\pi]$, the solutions are

$$S=\{\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6}\}.$$

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You start of correctly, but I do not know from where you get that $\cos x =1$. From the step \begin{equation} \cos x = 2 \sin x \cos x \end{equation} you can deduce that either $cos x = 0$ or $2 \sin x = 1$ (these are the only two possibilities for this equation. Can you see this?).

Now you can solve each of these cases. $\cos x = 0$ gives $ x = \pi/2 + n \pi$, where $n$ is an integer. The only solution that is in your range is $x = \pi/2$.

The second possibility $ \sin x = 1/2$ gives the solution $x = \pi/6 + 2 n \pi$ or $x = 5 \pi /6 + 2n\pi$ (to see this, draw the unit circle). The only solutions that are in your range are $x = \pi /6 $ and $x = 5 \pi / 6$.

Hence the solutions are $x = \pi/2 , x= \pi /6$ or $x = 5 \pi /6$.

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