Chapter 7.5, Problem 3E

To determine

To solve : the given equation $\{2{\cos }^2\theta +\sin \theta =1$ .

Answer to Problem 3E

The general solutions of $\theta$ for the equation $\{2{\cos }^2\theta +\sin \theta =1$ are

  $\theta =\frac{7\pi }{6}\text{+2n}\pi \text{ , }\theta \text{=}\frac{11\pi }{6}+2n\pi \text{ or }\theta =\frac{\pi }{2}+2n\pi$ .

Explanation of Solution

Given information:

The equation to be solved is $\{2{\cos }^2\theta +\sin \theta =1$ .

Calculation :

  $2{\cos }^2\theta +\sin \theta =1$.............. (1)

The trigonometric identity is $\{\begin{array}{l}{\cos }^2\theta +{\sin }^2\theta =1\\ {\cos }^2\theta =1-{\sin }^2\theta \end{array}$ .

Put the value of ${\cos }^2\theta$ in equation 1.

  $\begin{array}{l}2\left(1-{\sin }^2\theta \right)+\sin \theta =1\\ 2-2{\sin }^2\theta +\sin \theta =1\\ 2{\sin }^2\theta -\sin \theta -1=0\end{array}$

Put $\sin \theta =x$ .

  $\begin{array}{l}2x^2-x-1=0\\ 2x^2-2x+x-1=0\\ 2x\left(x-1\right)+1\left(x-1\right)=0\\ \left(2x+1\right)\left(x-1\right)=0\end{array}$

  $x=\frac{-1}{2}\text{ or }x=1$

Or

  $\sin \theta =\frac{-1}{2}\text{ or sin}\theta =1$

Now, there are two cases. Solve each case to get the solution of $\theta$ .

Case 1:

  $\sin \theta =\frac{-1}{2}$

Above equation is satisfied in the interval $\left[0,2\pi \right]$ when $\theta =\frac{7\pi }{6}\text{ or }\theta \text{=}\frac{11\pi }{6}$ .

Since, the period of sine function is $2\pi$ . Add $2\pi n$ to get the general solution of $\theta$ where $n$ is any integer.

  $\theta =\frac{7\pi }{6}\text{+2n}\pi \text{ or }\theta \text{=}\frac{11\pi }{6}+2n\pi$

Case 2:

  $\sin \theta =1$

Above equation is satisfied in the interval $\left[0,2\pi \right]$ when $\theta =\frac{\pi }{2}$ .

Since, the period of sine function is $2\pi$ . Add $2\pi n$ to get the general solution of $\theta$ where $n$ is any integer.

  $\theta =\frac{\pi }{2}+2n\pi$

Therefore, the general solutions of $\theta$ for the equation $\{2{\cos }^2\theta +\sin \theta =1$ are

  $\theta =\frac{7\pi }{6}\text{+2n}\pi \text{ , }\theta \text{=}\frac{11\pi }{6}+2n\pi \text{ or }\theta =\frac{\pi }{2}+2n\pi$ .