I can't find the intersections of (sqrt3)cos2x and sin2x. I know the basic method is to create a new equation from the two and obtain the x-intercepts.
Hi
If I have well understood you need to solve
$\displaystyle \sqrt{3} \cos(2x) = \sin(2x)$
$\displaystyle \sqrt{3} \cos(2x) - \sin(2x) = 0$
$\displaystyle \frac{\sqrt{3}}{2} \cos(2x) - \frac12 \sin(2x) = 0$
$\displaystyle \cos \left(\frac{\pi}{6}\right) \cos(2x) - \sin \left(\frac{\pi}{6}\right) \sin(2x) = 0$
$\displaystyle \cos \left(2x + \frac{\pi}{6}\right) = 0$
Can you get it from here ?
Hello, Stuck Man!
Find the intersections of: .$\displaystyle y \:=\:\sqrt{3}\cos2x\:\text{ and }\:y \:=\:\sin2x$
We have: .$\displaystyle \sin2x \;=\;\sqrt{3}\cos2x$
Then: .$\displaystyle \frac{\sin2x}{\cos2x} \:=\:\sqrt{3} \quad\Rightarrow\quad \tan2x \:=\:\sqrt{3}$
Hence: .$\displaystyle 2x \;=\;\frac{\pi}{3} + \pi n$
Therefore: .$\displaystyle x \;=\;\frac{\pi}{6} + \frac{\pi}{2}n$