Problem 54 - Algebra 2

Question

Find the complete solution in radians of each equation. $$ 4 \sin ^{2} \theta+1=4 \sin \theta $$

Step-by-Step Solution

Answer

The solution to the equation in radians is \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)

Step 1: Reformulate the Equation as Quadratic

Start by writing the trigonometric equation in the quadratic form, let \(x = \sin \theta\). Thus, the equation becomes: \(4x^2 - 4x + 1 = 0\)

Step 2: Apply the Quadratic Formula

Apply the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4, b = -4, c = 1\). Solving this yields \(x = \frac{4 \pm \sqrt{16 - 16}}{8} = \frac{1}{2}\)

Step 3: Solve for \(\theta\)

Now that we have \(x = \sin \theta = \frac{1}{2}\), we can find the angle \(\theta\) as \( \theta = \sin^{-1}(x) = \sin^{-1}(\frac{1}{2})\). We get \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\). It's worth noting that \(\sin (\pi - \theta) = \sin \theta\), which is the reason we have two solutions.

Key Concepts

Quadratic Formula

The quadratic formula is essential for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It's a formula that provides the solutions, or roots, of any quadratic equation. Here's what it looks like:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

To use the quadratic formula, identify and substitute the values of \(a\), \(b\), and \(c\) from your equation.
Then solve for \(x\) by resolving the plus-minus symbol \(\pm\), offering two possible solutions.
In our initial problem, the solving process starts by converting the trigonometric equation into a quadratic form, making it possible to apply the formula. The quadratic formula is our tool for unlocking the unknowns in mathematical equations by crunching numbers into tangible results.

Solving Equations

Solving equations is all about finding the unknown variable. In mathematics, equations set expressions equal and challenge you to find variable values satisfying this balance.
Each equation type demands a specific approach.

• Linear equations involve straight-line solutions.
• Quadratic equations involve squared terms, like in our exercise.

To solve, you often rearrange terms and use mathematical tools, such as quadratic formulas or inverse operations, to isolate and find the variable.
In this exercise, we converted the given trigonometric equation into a solvable quadratic form, using substitution (\(x = \sin \theta\)) to facilitate solving.

Sine Function

The sine function, \( \sin \theta \), is a fundamental trigonometric function representing the y-coordinate of a point on the unit circle corresponding to an angle \(\theta\). Typically, its values range from -1 to +1, oscillating smoothly as angles increase.
In trigonometry, the sine function is crucial for relating angles to right triangle sides, assessing oscillations, and describing wave motions.
For instance, in this specific problem where \(4 \sin^2 \theta + 1 = 4 \sin \theta\), we use the sine function to bridge angles and solve the equation, highlighting a practical application of sine in problem-solving.

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to determine angles when the function values, such as sine, cosine, or tangent, are known. These functions reverse the standard trigonometric roles.

• For sine, it is \( \sin^{-1} \) or arcsin.

When solving trigonometric equations, the inverse function helps find all possible angle solutions.
In our exercise, once we obtained \(x = \frac{1}{2}\), using the inverse sine function, \(\sin^{-1} (\frac{1}{2})\), furnished solutions for \(\theta\). Here, it delivers \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\), which represents how inverse functions rectify the direction of trigonometric functions, reaching back to angles.