Problem 53 - Algebra and Trigonometry

Question

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Solve \(2 \sin ^{2} \theta-\sin \theta+5=6\) for \(0 \leq \theta<2 \pi\)

Step-by-Step Solution

Answer

\( \theta = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6} \)

Step 1: Isolate the Trigonometric Equation

Start by isolating the trigonometric equation. We need to have the equation in the form of a standard quadratic equation.First, rewrite the given equation: \[ 2 \sin^2 \theta - \sin \theta + 5 = 6 \]Subtract 6 from both sides of the equation to set it to 0:\[ 2 \sin^2 \theta - \sin \theta + 5 - 6 = 0 \]Simplify it:\[ 2 \sin^2 \theta - \sin \theta - 1 = 0 \]

Step 2: Solve the Quadratic Equation

Now, solve the quadratic equation \(2 \sin^2 \theta - \sin \theta - 1 = 0\) using the quadratic formula.The quadratic formula is given by\[ \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In this case, \(a = 2\), \(b = -1\), and \(c = -1\).Substitute these values into the formula:\[ \sin \theta = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)} \]Simplify the expression:\[ \sin \theta = \frac{1 \pm \sqrt{1 + 8}}{4} \]\[ \sin \theta = \frac{1 \pm \sqrt{9}}{4} \]\[ \sin \theta = \frac{1 \pm 3}{4} \]

Step 3: Evaluate the Solutions

Now, evaluate the solutions from the quadratic equation.Case 1: \( \sin \theta = \frac{1 + 3}{4} = 1 \)Case 2: \( \sin \theta = \frac{1 - 3}{4} = -\frac{1}{2} \)This gives us the possible values:\( \sin \theta = 1 \) and \( \sin \theta = -\frac{1}{2} \)

Key Concepts

Quadratic Trigonometric Equations

Quadratic trigonometric equations are equations involving trigonometric functions where the highest degree of the variable is 2.
These types of equations combine concepts from both algebra and trigonometry, making them rather unique.
A typical quadratic trigonometric equation might look something like this:
\( a \, \sin^2 \theta + b \, \sin \, \theta + c = 0 \), where \( a, b, \) and \( c \) are constants.
To solve these equations, you often follow a similar process to solving quadratic equations in algebra.
First, it's essential to rewrite the problem in a standard quadratic form.
For instance, in the original exercise, we started with:
\[ 2 \, \sin^2 \, \theta - \sin \, \theta + 5 = 6 \]
We then isolated the quadratic part by subtracting 6 from both sides, resulting in:
\[ 2 \, \sin^2 \, \theta - \sin \, \theta - 1 = 0 \].
Once in standard form, we can proceed to solve it using various methods like factoring or the quadratic formula.

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \).
The formula itself is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
This solves for x, the variable in the equation.
Applying this to trigonometric equations is straightforward.
First, identify coefficients \( a, b, \) and \( c \).
In our example, the equation is:
\[ 2 \, \sin^2 \, \theta - \sin \, \theta - 1 = 0 \].
We identify that \( a = 2, b = -1, \) and \( c = -1 \).
Substituting these values into the quadratic formula gives:
\[ \sin \, \theta = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)} \].
Simplify to get:
\[ \sin \, \theta = \frac{1 \pm \sqrt{9}}{4} \].
This reduces further to:
\[ \sin \, \theta = \frac{1 + 3}{4} = 1 \] and \[ \sin \, \theta = \frac{1 - 3}{4} = -\frac{1}{2} \].
Thus, the quadratic formula helps us find trigonometric solutions efficiently.

Trigonometric Values

After solving quadratic trigonometric equations using the quadratic formula, you'll end up with specific trigonometric values.
In our case, we found:
\[ \sin \, \theta = 1 \] and \[ \sin \, \theta = -\frac{1}{2} \].
We now need to find all angles \( \theta \) in the given domain satisfying these values.
For \[ \sin \, \theta = 1 \], the sine function is 1 at \( \theta = \frac{\pi}{2} \text{ (90 degrees)} \).
For \[ \sin \, \theta = -\frac{1}{2} \], the sine function is -1/2 at \( \theta = \frac{7\pi}{6} \text{ (210 degrees)} \text{ and } \theta = \frac{11\pi}{6} \text{ (330 degrees)} \).
This means that within the domain \[ 0 \leq \theta < 2\pi \], these angles are the solutions.
When solving such equations, always remember to check the unit circle and reference angles to find all possible solutions within the given range. This is a crucial step in ensuring your answers are complete and accurate.