Chapter 9, Problem 1RE

For the following exercises, find all solutions exactly that exist on the interval [0, 2n).

1. ${\csc }^{2}t=3$

To determine

To find:

All exact solution on $\left[0,2\pi \right)$ for the given equation.

Answer to Problem 1RE

  $t=\boxed{\frac{49\pi }{250},\frac{201\pi }{250},\frac{299\pi }{250},\&\frac{451\pi }{250}}$

Explanation of Solution

Given information:

A trigonometric equation ${\csc }^{2}t=3$.

Formula Used:

    The sign of trigonometric ratio in different quadrants

  

Calculation:

Steps		Description
$\begin{array}{l}{\csc }^{2}t=3\\ \sqrt{{\csc }^{2}t}=\pm \sqrt{3}\\ \csc t=\pm \sqrt{3}\end{array}$		Using the square root property to solve
$\begin{array}{l}\csc t=\sqrt{3}\\ t={\csc }^{-1}\left(\sqrt{3}\right)\\ t=\frac{49\pi }{250},\frac{201\pi }{250}\end{array}$	$\begin{array}{l}\csc t=-\sqrt{3}\\ t={\csc }^{-1}\left(-\sqrt{3}\right)\\ t=\frac{299\pi }{250},\frac{451\pi }{250}\end{array}$	Solving for t

Conclusion:

The values of t which will make the given equation ${\csc }^{2}t=3$ true are $t=\boxed{\frac{49\pi }{250},\frac{201\pi }{250},\frac{299\pi }{250},\&\frac{451\pi }{250}}$.