Chapter 10, Problem 11CR

To determine

The solution of the equation $\cot \left(2\theta \right)=1$, where $0^{\circ }<\theta <{90}^{\circ }$

Answer to Problem 11CR

Solution:

The solution of equation $\cot \left(2\theta \right)=1$ on $\left(0,\frac{\pi }{2}\right)$ is $\frac{\pi }{8}$

Explanation of Solution

Given information:

The equation $\cot \left(2\theta \right)=1$, $0^{\circ }<\theta <{90}^{\circ }$

Here, $\cot \left(2\theta \right)=1$ implies that $\frac{1}{\tan \left(2\theta \right)}=1$

Thus, $\tan \left(2\theta \right)=1$

The tan is positive in I and III quadrant.

The argument is $2\theta$

Write the general formula that gives all the solutions of the equation $\tan \left(2\theta \right)=-1$, then list those are in the interval $\left(0,\frac{\pi }{2}\right)$

Since $\tan \left(\frac{\pi }{4}+k\pi \right)=1$ for any integer $k$ ,

  $\Rightarrow 2\theta =\frac{\pi }{4}+k\pi$

  $\Rightarrow \theta =\frac{\pi }{8}+\frac{k\pi }{2}$

If $k=0$, then $\theta =\frac{\pi }{8}+\frac{\left(0\right)\pi }{2}=\frac{\pi }{8}$

If $k=1$, then $\theta =\frac{\pi }{8}+\frac{\left(1\right)\pi }{2}=\frac{5\pi }{8}$

If $k=2$, then $\theta =\frac{\pi }{8}+\frac{\left(2\right)\pi }{2}=\frac{9\pi }{8}$

If $k=3$, then $\theta =\frac{\pi }{8}+\frac{\left(3\right)\pi }{2}=\frac{13\pi }{8}$

Therefore, the solution of equation $\cot \left(2\theta \right)=1$ on $\left(0,\frac{\pi }{2}\right)$ is $\frac{\pi }{8}$