Problem 63 - Algebra 2

Question

Find the complete solution of each equation. Express your answer in degrees. \(\sin ^{2} \theta+5 \sin \theta=0\)

Step-by-Step Solution

Answer

The equation \(\sin ^{2} \theta+5 \sin \theta=0\) has solutions \(\theta = 0°, 180°, 360°...\)

Step 1: Write the equation in standard form

The equation is already in the standard form of a quadratic equation, i.e., \(\sin ^{2} \theta+5 \sin \theta=0\)

Step 2: solve the quadratic equation

First, make the equation in the form \(ax^2 + bx = 0\), here \(x = \sin \theta\). So, you have \(x^2 + 5x = 0\). This equation can be solved by factoring and using the zero product property. So, \(x * (x+5) = 0\). This implies that either \(x = 0\) or \(x = -5)\.

Step 3: Solve for theta

\(\theta\) must be such that \(\sin(\theta) = x\). From step 2 we know that x can be 0 or -5. The first root \(x = 0\) implies that \(\sin(\theta) = 0\), which then implies \(\theta = 0°, 180°, 360°...\). The second root \(x = -5\) exceeds the possible values of \(\sin(\theta)\) since the range of \(\sin(\theta)\) is \([-1,1]\). So, \(x = -5\) is extraneous and not a valid solution.

Key Concepts

Quadratic Equations

A quadratic equation is a polynomial equation of degree two. It often takes the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants with \(a eq 0\). The equation provided, \(\sin^2 \theta + 5 \sin \theta = 0\), is a quadratic equation where the variable is \(\sin \theta\) rather than a typical \(x\).
To solve quadratic equations, you can use different methods:

• Factoring: If the quadratic can be factored into the product of two binomials, set each factor to zero and solve for the variable.
• Quadratic Formula: For more complex quadratics, the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) can find solutions.
• Completing the Square: This method rewrites the quadratic in a perfect square form.

In our solution, factoring was used because the equation format \(x^2 + 5x = 0\) easily allowed for it, revealing that \(x = 0\) or \(x = -5\), though not all values were valid.

Solution of Equations

Solving equations involves finding the values of the variable that make the equation true. For trigonometric equations like ours, we specifically solve for angles represented by \(\theta\). The general approach for solving these involves:

• Identifying the potential solutions of the transformed equation (here, \(\sin \theta = x\)).
• Applying the range or domain restrictions specific to trigonometric functions. The sine function, for example, only outputs values from \([-1, 1]\).
• Excluding extraneous solutions, which might appear naturally in the process of solving, such as complex numbers or values out of the function's range, like \(x = -5\) in our case.

Thus, solving \(\sin^2 \theta + 5 \sin \theta = 0\) involves factoring first, then testing the solutions against the possible range of the sine function to ensure they are valid.

Angle Measurement

In trigonometry, angles can be measured in degrees or radians, though degrees are often easier for basic exercises. A complete revolution around a circle equals \(360^\circ\) in degrees. When calculating solutions, like \(\theta\) in our scenario, your answers should typically fall within a single cycle of the sine wave for clarity. For instance, solving \(\sin(\theta) = 0\), gives multiple solutions since sine repeats every \(360^\circ\). The angles \(0^\circ\), \(180^\circ\), and \(360^\circ\) are all valid, known as co-terminal angles. These angles express cycles or repetitions crucial in periodic functions.
While solving trigonometric equations, it's important to remember to express answers in their most natural form, here in degrees, to make interpretation straightforward for common word problems and contexts.

Sine Function

The sine function, \(\sin(\theta)\), is a fundamental trigonometric function. It is periodic with a period of \(360^\circ\) and ranges from \(-1\) to \(1\). This is vital, as it restricts solutions to within that range.
The graph of the sine function has a wavelike pattern, intersecting the x-axis at angles that are multiples of \(180^\circ\), like \(0^\circ\), \(180^\circ\), and \(360^\circ\), which were part of our exercise's solution.
When we solve equations involving sine, like \(\sin(\theta) = 0\), these intersections guide us in determining where on the sine wave the function reaches zero. The repetitive nature of sine means there will be infinitely many solutions, but often only the smallest positive solutions are relevant. Understanding the behavior and range of the sine function ensures accurate and efficient problem-solving.