Chapter 7, Problem 26CT

To determine

The solution of the equation $-3\cos \left(\frac{\pi }{2}-\theta \right)=\tan \theta$ on $0\le \theta <2\pi$.

Answer to Problem 26CT

Solution:

The solutions of the equation $-3\cos \left(\frac{\pi }{2}-\theta \right)=\tan \theta$ on $0\le \theta <2\pi$ are $\theta =0,\pi ,1.911$ and $4.373$.

Explanation of Solution

Given information:

The equation is $-3\cos \left(\frac{\pi }{2}-\theta \right)=\tan \theta$

Explanation:

Consider the equation $-3\cos \left(\frac{\pi }{2}-\theta \right)=\tan \theta$.

By using $\cos \left(\frac{\pi }{2}-\theta \right)=\sin \theta$,

$\Rightarrow -3\sin \theta =\tan \theta$

Adding $3\sin \theta$ from both sides,

$\Rightarrow \tan \theta +3\sin \theta =0$

$\Rightarrow \frac{\sin \theta }{\cos \theta }+3\sin \theta =0$

$\Rightarrow \sin \theta \left(\frac{1}{\cos \theta }+3\right)=0$

By using zero product property,

$\Rightarrow \sin \theta =0$ or $\frac{1}{\cos \theta }+3=0$

$\Rightarrow \sin \theta =0$ or $\cos \theta =-\frac{1}{3}$

The solutions of $\sin \theta =0$ are $\theta =0$ or $\pi$.

Let $\cos \theta =-\frac{1}{3}$ which implies that $\theta ={\cos }^{-1}\left(-\frac{1}{3}\right)$

The cosine function is negative in $II$ and $III$ quadrant.

The solutions of $\cos \theta =-\frac{1}{3}$ are $\theta =\pi -{\cos }^{-1}\frac{1}{3}$ and $\pi +{\cos }^{-1}\frac{1}{3}$.

This implies that $\theta =1.91063323$ and $\theta =4.37255207$.

That is $\theta \approx 1.911$ or $\theta \approx 4.373$.

Therefore, the solution of the equation $-3\cos \left(\frac{\pi }{2}-\theta \right)=\tan \theta$ on $0\le \theta <2\pi$ are $\theta =0,\pi ,1.911$, and $4.373$.