Chapter 7.4, Problem 6E

To determine

To find: The solution of the equation $\sin \theta =-\frac{\sqrt{2}}{2}$ .

Answer to Problem 6E

The solutions of the equation $\sin \theta =-\frac{\sqrt{2}}{2}$ are $\frac{5\pi }{4}+2k\pi$ and $\frac{7\pi }{4}+2k\pi$ .

Explanation of Solution

Given information:

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

Calculation:

Since, sine has period $2\pi$ . So first find the solutions in $\left[0,2\pi \right)$ . The function sine is negative in the third and the fourth quadrants.

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

  $\begin{array}{c}\theta =\pi +\frac{\pi }{4},2\pi -\frac{\pi }{4}\\ =\frac{5\pi }{4},\frac{7\pi }{4}\end{array}$

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

So, the general solution of the equation is given by

  $\theta =\frac{5\pi }{4}+2k\pi ,\frac{7\pi }{4}+2k\pi$

Therefore, the solutions of the equation $\sin \theta =-\frac{\sqrt{2}}{2}$ are $\frac{5\pi }{4}+2k\pi$ and $\frac{7\pi }{4}+2k\pi$ .