Solve the equation sin(x+pi/6)=2cosx in the range of 0<x<2pi?

1 Answer
Feb 22, 2018

x=\frac{\pi }{3}" "\color(blue){and}" " x=\frac{4\pi }{3}

Explanation:

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}