Question

Solve the following equations.
$sin(\pi -6x)+\sqrt{3}sin(\frac{\pi }{2}+6x)=\sqrt{3}.$

Answer 1

$sin(\pi -6x)+\sqrt{3}sin(\frac{\pi }{2}+6x)=\sqrt{3}$

$\Rightarrow sin\left(6x\right)+\sqrt{3}cos6x=\sqrt{3}\{\because sin(\pi -\theta )=sin\theta \}$

$\{\because sin(\frac{\pi }{2}+\theta )=cos\theta \}$

$\Rightarrow \frac{1}{2}sin6x+\frac{\sqrt{3}}{2}cos6x=\frac{\sqrt{3}}{2}...$ [ Dividing both sides by 2]

$\Rightarrow \frac{\sqrt{3}}{2}cos6x+\frac{1}{2}sin6x=\frac{\sqrt{3}}{2}$

$\Rightarrow cos6x.cos\left(\frac{\pi }{6}\right)+sin6x.sin\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\{\because cos\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}andsin\left(\frac{\pi }{6}\right)=\frac{1}{2}\}$

$\Rightarrow cos(6x-\frac{\pi }{6})=cos\left(\frac{\pi }{6}\right)\{\because cos(A-B)=cosA.cosB+sinA.sinB\because cos\left(\frac{\pi }{6}\right)=\sqrt{3}/2\}$

$\Rightarrow$ Now we know General solution for

$cos\theta =cos\alpha \Rightarrow \{\theta =\alpha +2n\pi ,nϵZ\}Z\to$ intergers $(0,\pm 1,\pm 2,\pm 3,\pm )$

Here $\alpha =\pi /6,$ for

$\Rightarrow cos(6x-\frac{\pi }{6})=cos\pi /6$

The general solution will be

$6x-\frac{\pi }{6}=\frac{\pi }{6}+2n\pi ,nϵZ$

$6x=\frac{\pi }{3}+2n\pi ,nϵZ$

$x=\frac{\pi }{18}+n\frac{\pi }{3},nϵZ$