Chapter 7.4, Problem 27E

To determine

To find: The solution of the equation $\sqrt{2}\sin \theta +1=0$ .

Answer to Problem 27E

The solutions of the equation $\sqrt{2}\sin \theta +1=0$ are $\theta =\frac{5\pi }{4}+2k\pi ,\frac{7\pi }{4}+2k\pi$ .

Explanation of Solution

Given information:

The given equation is $\sqrt{2}\sin \theta +1=0$ .

Calculation:

Simplify the given equation.

  $\begin{array}{l}\sqrt{2}\sin \theta +1=0\\ \sin \theta =\frac{-1}{\sqrt{2}}\end{array}$

As 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

  $\begin{array}{c}\theta ={\sin }^{-1}\left(-\frac{1}{\sqrt{2}}\right)\\ =\frac{5\pi }{4},\frac{7\pi }{4}\end{array}$

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 $\sqrt{2}\sin \theta +1=0$ are $\theta =\frac{5\pi }{4}+2k\pi ,\frac{7\pi }{4}+2k\pi$ .