# Solve this equation for x \in (0,\frac{\pi}{2}) \cos(2x)\cos(x-\frac{\pi}{6})=\sin(2x)\sin(\frac{\pi}{6}-x) I gave a try

Solve this equation for $x\in \left(0,\frac{\pi }{2}\right)$
$\mathrm{cos}\left(2x\right)\mathrm{cos}\left(x-\frac{\pi }{6}\right)=\mathrm{sin}\left(2x\right)\mathrm{sin}\left(\frac{\pi }{6}-x\right)$
I gave a try using the $\mathrm{cos}\left(a-b\right)$ and $\mathrm{sin}\left(a-b\right)$ formulas, but it seems the problem complicated a little bit more.
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temzej9
$\mathrm{cos}\left(2x\right)\mathrm{cos}\left(\frac{\pi }{6}-x\right)-\mathrm{sin}\left(2x\right)\mathrm{sin}\left(\frac{\pi }{6}-x\right)=0$
$⇒\mathrm{cos}\left(2x+\frac{\pi }{6}-x\right)=0$
(using $\mathrm{cos}\left(a+b\right)=\mathrm{cos}a\mathrm{cos}b-\mathrm{sin}a\mathrm{sin}b$)
$⇒x+\frac{\pi }{6}=\frac{\pi }{2}$
$⇒x=\frac{\pi }{3}$
###### Not exactly what you’re looking for?
RizerMix
It is not complicated, the equation become $\mathrm{cos}\left(2x\right)\mathrm{cos}\left(x-\frac{\pi }{6}\right)+\mathrm{sin}\left(2x\right)\mathrm{sin}\left(x-\frac{\pi }{6}\right)=0$ that is equivalent to $\mathrm{cos}\left(x+\frac{\pi }{6}\right)=0$ Thus the solution is $x+\frac{\pi }{6}=\frac{\pi }{2}+k\pi$ $x=\frac{\pi }{3}+k\pi$
###### Not exactly what you’re looking for?
user_27qwe
The equation is $\mathrm{cos}\left(2x\right)\mathrm{cos}\left(x-\frac{\pi }{6}\right)+\mathrm{sin}\left(2x\right)\mathrm{sin}\left(x-\frac{\pi }{6}=0$. This is same as $\mathrm{cos}\left(2x-\left(x-\frac{\pi }{6}\right)\right)=0$ or $\mathrm{cos}\left(x+\frac{\pi }{6}\right)=0$. So $x+\frac{\pi }{6}=\frac{\left(2n+1\right)\pi }{2}$ for some integer n. For $x\in \left(0,\frac{\pi }{2}\right)$ we must have n=0 so $x=\frac{\pi }{3}$