Chapter 7.4, Problem 20E

To determine

To find: The solution of the equation $\sin \theta =-\frac{\sqrt{3}}{2}$ and list six specific solutions.

Answer to Problem 20E

The solutions of the equation $\sin \theta =-\frac{\sqrt{3}}{2}$ are $\theta =\frac{4\pi }{3}+2k\pi ,\frac{5\pi }{3}+2k\pi$ and six specific solutions are $-\frac{2\pi }{3},-\frac{\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3},\frac{10\pi }{3},\frac{11\pi }{3}$ .

Explanation of Solution

Given information:

The given equation is $\sin \theta =-\frac{\sqrt{3}}{2}$ .

Calculation:

As sine has period $2\pi$ , So first find the solutions in $\left[0,2\pi \right)$ .

As the given equation is $\sin \theta =-\frac{\sqrt{3}}{2}$ .

Function sine is negative in third and fourth Quadrants.

Therefore, the solution to the equation in $\left[0,2\pi \right)$ are

  $\begin{array}{c}\theta ={\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\\ =\frac{4\pi }{3},\frac{5\pi }{3}\end{array}$

As cosine repeats its values after every $2\pi$ units.

So, The general solution of the equation is given by

  $\theta =\frac{4\pi }{3}+2k\pi ,\frac{5\pi }{3}+2k\pi$ .

Now substitute $k=-1,0,1$ in the above solutions

  $\begin{array}{c}\theta =\frac{4\pi }{3}-2\pi ,\frac{5\pi }{3}-2\pi ,\frac{4\pi }{3},\frac{5\pi }{3},\frac{4\pi }{3}+2\pi ,\frac{5\pi }{3}+2\pi \\ =-\frac{2\pi }{3},-\frac{\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3},\frac{10\pi }{3},\frac{11\pi }{3}\end{array}$

Therefore the six specific solutions of the equation $\sin \theta =-\frac{\sqrt{3}}{2}$ are $-\frac{2\pi }{3},-\frac{\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3},\frac{10\pi }{3},\frac{11\pi }{3}$ .

Therefore, the solutions of the equation $\sin \theta =-\frac{\sqrt{3}}{2}$ are $\theta =\frac{4\pi }{3}+2k\pi ,\frac{5\pi }{3}+2k\pi$ .