Chapter 5.4, Problem 72E

To determine

All solutions of the below given equation in the interval $\left[0,2\pi \right)$ .

  $\sin \left(x+\frac{\pi }{6}\right)-\sin \left(x-\frac{7\pi }{6}\right)=\frac{\sqrt{3}}{2}$

Answer to Problem 72E

  $x=\frac{\pi }{6},\frac{5\pi }{6}$

Explanation of Solution

Given Information:

The given equation is:

  $\sin \left(x+\frac{\pi }{6}\right)-\sin \left(x-\frac{7\pi }{6}\right)=\frac{\sqrt{3}}{2}$

Formula Used:

Using formulas:

  $\mathrm{sinC}-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$

Calculation:

  $\sin \left(x+\frac{\pi }{6}\right)-\sin \left(x-\frac{7\pi }{6}\right)=\frac{\sqrt{3}}{2}$

Using the difference of cosine formula, we get:

  $\begin{array}{l}\Rightarrow 2\cos \left(\frac{x+\frac{\pi }{6}+x-\frac{7\pi }{6}}{2}\right)\sin \left(\frac{x+\frac{\pi }{6}-x+\frac{7\pi }{6}}{2}\right)=\frac{\sqrt{3}}{2}\\ \Rightarrow 2\cos \left(\frac{2x-\frac{6\pi }{6}}{2}\right)\sin \left(\frac{\frac{8\pi }{6}}{2}\right)=\frac{\sqrt{3}}{2}\\ \Rightarrow 2\cos \left(\frac{2x-\pi }{2}\right)\sin \left(\frac{\frac{4\pi }{3}}{2}\right)=\frac{\sqrt{3}}{2}\\ \Rightarrow 2\cos \left(x-\frac{\pi }{2}\right)\sin \left(\frac{2\pi }{3}\right)=\frac{\sqrt{3}}{2}\end{array}$

We know that:

  $\sin \left(\frac{2\pi }{3}\right)=\frac{\sqrt{3}}{2}$

Plugging the above value, we get:

  $\begin{array}{l}\Rightarrow 2\cos \left(x-\frac{\pi }{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{2}\\ \Rightarrow \sqrt{3}\cos \left(x-\frac{\pi }{2}\right)=\frac{\sqrt{3}}{2}\end{array}$

Isolating cos term, we get:

  $\begin{array}{l}\Rightarrow \cos \left(x-\frac{\pi }{2}\right)=\frac{\sqrt{3}}{2\sqrt{3}}\\ \Rightarrow \cos \left(x-\frac{\pi }{2}\right)=\frac{1}{2}\end{array}$

Taking sine inverse both sides, we get:

  $\Rightarrow x-\frac{\pi }{2}={\cos }^{-1}\left(\frac{1}{2}\right)$

We know that

  $\begin{array}{l}{\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}+2n\pi ,\frac{5\pi }{3}+2n\pi \\ \text{where }n\text{ is an integer}\end{array}$

Hence we get:

  $\begin{array}{l}\Rightarrow x-\frac{\pi }{2}=\frac{\pi }{3}+2n\pi \text{ or }\Rightarrow x-\frac{\pi }{2}=\frac{5\pi }{3}+2n\pi \\ \Rightarrow x=\frac{\pi }{3}+2n\pi +\frac{\pi }{2}\text{ or }\Rightarrow x=\frac{5\pi }{3}+2n\pi +\frac{\pi }{2}\\ \Rightarrow x=\frac{5\pi }{6}+2n\pi \text{ or }\Rightarrow x=\frac{13\pi }{6}+2n\pi \end{array}$

Hence the possible values of x are:

  $\Rightarrow x=-\frac{7\pi }{6},\frac{\pi }{6},\frac{5\pi }{6},\frac{13\pi }{6},\frac{19\pi }{6},\frac{25\pi }{6},\mathrm{.....}$

Possible values of x in the interval $\left[0,2\pi \right)$ are:

  $\Rightarrow x=\frac{\pi }{6},\frac{5\pi }{6}$