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:
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.
Best Answer
$$ \cos(x)=\sin(2x)$$
$$ \Leftrightarrow 2\sin(x)\cos(x)-\cos(x)=0$$
$$ \Leftrightarrow \cos(x)(2\sin(x)-1)=0$$
$$ \Leftrightarrow \cos(x)=0 ~~~\sin(x)=\frac{1}{2}$$
$$ \Leftrightarrow x=\pm\frac{\pi}{2}+2\pi k ~~~ x=\pi-\frac{\pi}{6}+2k\pi $$
where $k \in Z$
Picture of intersections: