Problem 61 - Precalculus: Functions and Graphs

Question

Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$2 \tan t \csc t+2 \csc t+\tan t+1=0$$

Step-by-Step Solution

Answer

The solutions are \(t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\).

Step 1: Simplify the equation

Start by rewriting the original equation: \(2 \tan t \csc t + 2 \csc t + \tan t + 1 = 0\). Factor out \(\csc t\) from the first two terms to get: \(\csc t(2 \tan t + 2) + \tan t + 1 = 0\).

Step 2: Factor further

Notice that \(\csc t = \frac{1}{\sin t}\) and \(\tan t = \frac{\sin t}{\cos t}\). Substituting these identities, the equation becomes \(\frac{2 \sin t}{\sin t \cos t} + \frac{2}{\sin t} + \frac{\sin t}{\cos t} + 1 = 0\). Simplify the fraction where possible.

Step 3: Multiply through by \(\sin t \cos t\)

To eliminate fractions, multiply every term by \(\sin t \cos t\): \(2 \cos t + 2 \cos t \sin t + \sin^2 t + \sin t \cos t = 0\).

Step 4: Rearrange and simplify

Combine like terms: \(2 \cos t + \sin t (2 \cos t + \sin t) = 0\). This suggests two possibilities: \(2 \cos t = 0\) or \(\sin t (2 \cos t + \sin t) = 0\).

Step 5: Solve the first part \(2 \cos t = 0\)

Solve for \(t\) when \(\cos t = 0\). This occurs at \(t = \frac{\pi}{2}\) and \(t = \frac{3\pi}{2}\) within the interval \([0, 2\pi)\).

Step 6: Solve the second part \(\sin t (2 \cos t + \sin t) = 0\)

This gives either \(\sin t = 0\) or \(2 \cos t + \sin t = 0\). From \(\sin t = 0\), \(t = 0\) or \(t = \pi\).

Step 7: Solve \(2 \cos t + \sin t = 0\)

Rearrange to \(2 \cos t = -\sin t\). Dividing through by \(\cos t\), we get \(2 = -\tan t\) or \(\tan t = -2\). Solve this within the interval \([0, 2\pi)\) to find \(t = \tan^{-1}(-2)\), considering calculator values and symmetry.

Key Concepts

Trigonometric Identities

Trigonometric identities are crucial for simplifying and manipulating trigonometric equations. In this exercise, we encounter terms like \( \tan t \) and \( \csc t \). These can be expressed using fundamental trigonometric identities:
- \( \tan t = \frac{\sin t}{\cos t} \)- \( \csc t = \frac{1}{\sin t} \)
These identities allow us to transform the equation into a form that is easier to work with by eliminating the trigonometric functions, which can help in factoring and simplifying the equation. By substituting these identities, we substitute complex trigonometric terms with basic algebraic expressions. Hence, it simplifies the given equation to a form where factoring becomes more evident.

Interval Solutions

Finding solutions within a specified interval requires considering the periodic nature of trigonometric functions. The interval \([0, 2\pi)\) represents one full cycle of the sine and cosine functions. This means any solution to the trigonometric equation must fall within this interval.
When solving trigonometric equations, it helps to visualize the unit circle. Solutions often correspond to angles on this circle. For instance:- If \( \cos t = 0 \), the angles are \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \).- If \( \sin t = 0 \), the angles are 0 and \( \pi \).
It's fundamental to find all possible angle solutions within the interval and verify them. Calculators can assist in finding the principal values of inverse trigonometric functions but remember to check the periodic nature for complete solutions.

Factoring Trigonometric Expressions

Factoring is a powerful technique to solve equations, including trigonometric ones. In this exercise, the equation \(2 \tan t \csc t + 2 \csc t + \tan t + 1 = 0\) was simplified by factoring out common terms initially and then rearranging into simpler parts.
The process of factoring begins by identifying common factors in the equation, such as \( \csc t \) in the first terms. Factoring transforms the equation into \( \csc t(2 \tan t + 2) + \tan t + 1 = 0\).This step reveals hidden relationships between terms, leading to separate factor equations that are easier to solve.
Finally, solve each factor separately to find all possible solutions, ensuring none are overlooked within the provided interval.