sin(x+\frac{\pi }{6})=2cos(x)
Manipulate the left hand side by using the identity:
\sin (s+t)=\cos (s)\sin (t)+\cos (t)\sin (s)
\cos (x)\sin (\frac{\pi }{6})+\cos (\frac{\pi }{6})\sin (x)=2cos(x)
Put the values \sin (\frac{\pi }{6})=\frac{1}{2} \color(red){and} \cos (\frac{\pi }{6})=\frac{\sqrt{3}}{2}
\cos (x)\sin (\frac{\pi }{6})+\cos (\frac{\pi }{6})\sin (x)=2cos(x)
\frac{1}{2}\cos (x)+\frac{\sqrt{3}}{2}\sin (x)=2cos(x)
Subtract 2cos(x) from both sides and then simplify to get:
\sqrt{3}\sin (x)-3\cos (x)=0
Dividing both sides with cos(x) will give us:
\frac{\sqrt{3}\sin (x)-3\cos (x)}{\cos (x)}=\frac{0}{\cos (x)}
Simplify:
\tan (x)=\sqrt{3}
General solutions for the above equation are x=\frac{\pi }{3}+\pi n
Solutions within the specified range are:
x=\frac{\pi }{3}" "\color(blue){and}" " x=\frac{4\pi }{3}