Chapter 7.5, Problem 9E

To determine

To solve : the given equation $\{\cos 2\theta =3\sin \theta -1$ .

Answer to Problem 9E

The general solutions of $\theta$ for the equation $\{\cos 2\theta =3\sin \theta -1$ are

  $\theta =\frac{\pi }{6}+2\pi n\text{ or }\theta =\frac{5\pi }{6}+2\pi n$ .

Explanation of Solution

Given information:

The equation to be solved is $\{\cos 2\theta =3\sin \theta -1$ .

Calculation :

  $\cos 2\theta =3\sin \theta -1$........... (1)

The trigonometric identity is $\{\cos 2\theta =1-2{\sin }^2\theta$ .

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

  $\begin{array}{l}\left(1-2{\sin }^2\theta \right)=3\sin \theta -1\\ 3\sin \theta -1-1+2{\sin }^2\theta =0\\ 2{\sin }^2\theta +3\sin \theta -2=0\end{array}$

Put $\sin \theta =x$ .

  $2x^2+3x-2=0$

Factorize the above equation.

  $\begin{array}{l}2x^2+4x-x-2=0\\ 2x\left(x+2\right)-1\left(x+2\right)=0\\ \left(2x-1\right)\left(x+2\right)=0\\ x=\frac{1}{2}\text{ or }x=-2\end{array}$

Or

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

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 $\theta =\frac{\pi }{6}\text{ or }\theta \text{=}\frac{5\pi }{6}$ .

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

  $\theta =\frac{\pi }{6}+2\pi n\text{ or }\theta =\frac{5\pi }{6}+2\pi n$

Case 2:

  $\sin \theta =-2$

Since, $-1\le \sin \theta \le 1$ . So, $\sin \theta =-2$ is invalid.

Therefore, the general solutions of $\theta$ for the equation $\{\cos 2\theta =3\sin \theta -1$ are

  $\theta =\frac{\pi }{6}+2\pi n\text{ or }\theta =\frac{5\pi }{6}+2\pi n$ .