Question

Nora is required to solve the following trigonometric equation. \(9 \sin ^{2} \theta+12 \sin \theta+4=0, \theta \in\left[0^{\circ}, 360^{\circ}\right)\) Nora did the work shown below. Examine her work carefully. Identify any errors. Rewrite the solution, making any changes necessary for it to be correct. \(9 \sin ^{2} \theta+12 \sin \theta+4=0\) \((3 \sin \theta+2)^{2}=0\) $$ 3 \sin \theta+2=0 $$ Therefore. \(\sin \theta=-\frac{2}{3}\) Use a colculator. \(\sin ^{-1}\left(-\frac{2}{3}\right)=-41.8103149\) So, the reference ongle is 41.8 , to the neorest tenth of a degree Sine is negotive in quodrants II ond III. The solution in quadront II is \(180^{\circ}-41.8^{\circ}=138.2\) The solution in quadrant III is \(180^{\circ}+41.8=221.8\) Therefore, \(\theta=138.2^{\circ}\) ond \(\theta=221.8\), to the neorest tenth of a degree.

Short answer

\( \theta = 221.8\{}\degree \text{ and } 318.2\{}\degree \)

Step-by-step solution

- Identify the quadratic equation

Start by identifying the given quadratic trigonometric equation: \text{ }\(9 \sin^{2} \theta + 12 \sin \theta + 4 = 0\)

- Factor the quadratic equation

To factor the equation, notice that it can be rewritten as a perfect square: \text{ }\((3 \sin \theta + 2)^{2} = 0\)

- Solve the factor

Since we have a squared term equal to zero, set the inner expression to zero: \text{ }\(3 \sin \theta + 2 = 0\)

- Solve for \sin \theta

Isolate \sin \theta: \text{ }\(3 \sin \theta = -2\) \text{ }\(\sin \theta = -\frac{2}{3}\)

- Find the reference angle

Calculate the reference angle using the inverse sine function: \text{ }\(\theta_{ref} = \sin^{-1}\left(\-\frac{2}{3}\right)\approx -41.8103\{}\degree\) Since we need the positive equivalent, the reference angle = \( 41.8\{}\degree\)

- Determine the angles in the specified range

Since \sin is negative in quadrants III and IV, we calculate: In quadrant III: \text{ }\(180\{}\degree + 41.8\{}\degree = 221.8\{}\degree\) In quadrant IV: \text{ }\(360\{}\degree - 41.8\{}\degree = 318.2\{}\degree\)

Key concepts

Trigonometry

Understanding trigonometry is crucial when solving equations involving angles. Trigonometry deals with the relationships between the angles and sides of triangles, especially right triangles. At its core, trigonometry uses functions such as sine, cosine, and tangent to relate angles to side lengths. In our exercise, we are working with the sine function. Knowing the basic properties and graphs of these functions helps us solve equations. Using tools like the unit circle, we can find angles that correspond to specific sine values. Additionally, the behavior of these functions within different quadrants of the unit circle is essential to solve trigonometric equations accurately.

Quadratic Equations

Quadratic equations are polynomial equations of degree 2. They typically appear in the form ax^2 + bx + c = 0. Solving these involves finding the values of x (or in trigonometry, the angle θ) that satisfy the equation.
In our particular exercise, the goal is to solve for θ where the equation is given as 9sin^2(θ) + 12sin(θ) + 4 = 0. This resembles a standard quadratic equation where the variable is sin(θ).
Factoring is one common method to solve quadratic equations if the quadratic can be written as a product of binomials. Here, we factored the trigonometric quadratic to (3sin(θ) + 2)^2 = 0. Solving this factored form helps us find the solutions for sin(θ).

Reference Angle

A reference angle is the acute angle formed by the terminal side of an angle and the x-axis. Reference angles are used to simplify the process of finding trigonometric function values for larger angles by relating them to function values in the first quadrant.
When solving trigonometric equations, once we find the inverse sine value (or any trig function), the resulting angle might need to be adjusted into the correct quadrant.
For instance, in our exercise, after finding that sin(θ) = -2/3, we find its inverse sine, which gives us an angle in the fourth quadrant. The reference angle helps us determine the equivalent angles in the other quadrants where sine is negative (quadrants III and IV). This way, we account for all possible solutions within the given interval.

Inverse Sine Function

The inverse sine function, denoted as sin^(-1) or arcsin, is used to find an angle whose sine value is given. It essentially reverses the sine function, making it an invaluable tool in trigonometry for solving equations.
For example, if sin(θ) = -2/3, then θ = sin^(-1)(-2/3). Calculating this gives an angle in radians or degrees that represents the initial solution but may fall outside the desired range or quadrant.
In our exercise, sin^(-1)(-2/3) gives an angle of -41.81 degrees, which is not within the standard range of 0° to 360°. By interpreting this using reference angles, we convert this to a more usable angle measure that adheres to the given interval. Thus, understanding how to apply and interpret the inverse sine function is a critical step in solving trigonometric equations accurately.