Problem 162 - Calculus Volume 1

Question

Solve the trigonometric equations on the interval \(0 \leq \theta<2 \pi\). \(\csc ^{2} \theta+2 \csc \theta+1=0\)

Step-by-Step Solution

Answer

\( \theta = \frac{3\pi}{2} \) is the solution.

Step 1: Rewrite in terms of \( ext{cosecant} \)

The given equation is \( \csc^2 \theta + 2 \csc \theta + 1 = 0 \). Let's substitute \( x = \csc \theta \) to simplify the equation. This gives us:\[ x^2 + 2x + 1 = 0 \]

Step 2: Factor the quadratic equation

The equation \( x^2 + 2x + 1 = 0 \) can be factored as:\[ (x + 1)(x + 1) = 0 \]So the solutions are \( x = -1 \) (a repeated root).

Step 3: Translate solutions back to \( \theta \)

Since we set \( x = \csc \theta \), and \( \csc \theta = -1 \), this means:\[ \frac{1}{\sin \theta} = -1 \]Solving this gives us \( \sin \theta = -1 \).

Step 4: Find solutions for \( \theta \) within the interval

\( \sin \theta = -1 \) at \( \theta = \frac{3\pi}{2} \) within the interval \( 0 \leq \theta < 2\pi \).

Key Concepts

Cosecant Function

The **cosecant function** is one of the six main trigonometric functions and is denoted by \( \csc \theta \). It is the reciprocal of the sine function, so \( \csc \theta = \frac{1}{\sin \theta} \). This function is only defined where the sine function is not zero, as division by zero is undefined.

In simpler terms, whenever you need to find \( \csc \theta \), think about flipping the sine of the angle. If \( \sin \theta \) is a fraction like \( \frac{a}{b} \), then \( \csc \theta = \frac{b}{a} \). Knowing that, let's delve into the importance of replacing \( \csc \theta \) with a variable \( x \), as performed in the exercise:

• This helps us see the equation as a quadratic in \( x \), making it easier to solve.
• We can then translate the solution for \( x \) back into terms of \( \theta \) to solve the trigonometric equation.

Quadratic Equations

**Quadratic equations** form the basis for many algebraic problems and appear often in trigonometry when working with reciprocal functions. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \).

Solving a quadratic equation can be done by factoring, completing the square, or applying the quadratic formula. In our exercise, since Alice is the author, the equation \( x^2 + 2x + 1 = 0 \) was solved by factoring:

• The equation factors into \((x + 1)(x + 1) = 0\).
• This implies that \( x = -1 \) is a repeated root, also known as a double root.

Factoring simplifies solving, as it breaks down the quadratic into simpler, solvable parts. When the quadratic equation is solved, the solutions for \( x \) are used to find the angles \( \theta \) in the trigonometric equation.

Sine Function

The **sine function** is fundamental in trigonometry, denoted by \( \sin \theta \). It relates to the ratio of the length of the opposite side to the hypotenuse in right-angled triangles.

In the exercise, after determining \( x = \csc \theta = -1 \), it's translated to find \( \sin \theta = -1 \). This step bridges the quadratic solution back to trigonometry:

• By rearranging, \( \sin \theta = \frac{1}{\csc \theta} \).
• This means \( \sin \theta = -1 \) when \( \csc \theta = -1 \).

Finally, understanding when \( \sin \theta = -1 \) is crucial. This particular value occurs at an angle of \( \theta = \frac{3\pi}{2} \) within the given interval \(0 \leq \theta < 2\pi \). Knowing these trigonometric values by heart, or their unit circle positions, can simplify solving such equations.