Problem 47 - Trigonometry

Question

\(2 \sin ^{2} \theta-3 \sin \theta=-1\)

Step-by-Step Solution

Answer

The solutions are \(\theta = \frac{\pi}{2}, \frac{\pi}{6}, 5\frac{\pi}{6}\) plus any multiple of \(2\pi\).

Step 1: Transform the Trigonometric Equation

First, let's rewrite the given equation in terms of a standard quadratic form. Start with \(2 \sin^2 \theta - 3 \sin \theta = -1\), and add 1 to both sides to get: \(2 \sin^2 \theta - 3 \sin \theta + 1 = 0\).

Step 2: Substitution for Simplification

Let \(x = \sin \theta\). This changes the equation to: \(2x^2 - 3x + 1 = 0\).

Step 3: Solve the Quadratic Equation

We will solve the quadratic equation \(2x^2 - 3x + 1 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 2\), \(b = -3\), and \(c = 1\).

Step 4: Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\). Substituting the values, \((-3)^2 - 4(2)(1) = 9 - 8 = 1\). Since the discriminant is positive, there are two real solutions.

Step 5: Use the Quadratic Formula

Substitute the calculated values into the quadratic formula: \(x = \frac{-(-3) \pm \sqrt{1}}{2 \times 2}\), which simplifies to \(x = \frac{3 \pm 1}{4}\).

Step 6: Find the Values of x

Calculate the two potential solutions for \(x\):- \(x = \frac{3 + 1}{4} = 1\)- \(x = \frac{3 - 1}{4} = \frac{1}{2}\)

Step 7: Determine the Trigonometric Solutions

Now revert back to \(\sin \theta = x\) with possible values \(\sin \theta = 1\) or \(\frac{1}{2}\). - For \(\sin \theta = 1\), \(\theta = \frac{\pi}{2} + 2n\pi\), where \(n\) is an integer,- For \(\sin \theta = \frac{1}{2}\), \(\theta = \frac{\pi}{6} + 2n\pi\) or \(5\frac{\pi}{6} + 2n\pi\), where \(n\) is an integer.

Key Concepts

Understanding Trigonometric Identities

Trigonometric identities are essential tools in simplifying and solving equations that involve trigonometric functions like sine, cosine, and tangent. They allow us to transform expressions into more manageable forms.
One of the fundamental identities is the Pythagorean identity: \(\sin^2 \theta + \cos^2 \theta = 1\).

• This identity relates sine and cosine, serving as a basis for many transformations.
• By rearranging it, you can express \(\sin^2 \theta\) as \(1 - \cos^2 \theta\) or \(\cos^2 \theta\) as \(1 - \sin^2 \theta\).

These identities are also useful in combining and converting between trigonometric functions. Applying them wisely can simplify an equation like \(2 \sin^2 \theta - 3 \sin \theta = -1\) into a standard quadratic form for easier solving.

Exploring Sinusoidal Functions

Sinusoidal functions, represented by sine and cosine waves, play a crucial role in modeling periodic phenomena. These functions are characterized by their amplitude, period, and phase.

• Amplitude: This is the peak value of the wave from its rest position. For \(\sin \theta\), the standard amplitude is 1, unless modified by a coefficient.
• Period: Defined as the distance over which the function repeats itself. For sine and cosine, the standard period is \(2\pi\).
• Phase Shift: This determines where the wave starts. Adding or subtracting a constant to the angle inside the function shifts it horizontally.

By understanding these properties, you can determine specific values and behavior of the waveforms at any given point, critical for solving trigonometric equations using solutions like \(\sin \theta = \frac{1}{2}\).

Solving Quadratic Equations Using the Quadratic Formula

Quadratic equations can often appear in trigonometry, especially when solving for angles using functions like sine and cosine. The quadratic formula is a powerful method for finding solutions to these equations:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

• Coefficients: In the general quadratic equation \(ax^2 + bx + c = 0\), \(a\), \(b\), and \(c\) are the coefficients.
• Discriminant: The expression \(b^2 - 4ac\) is known as the discriminant. The value of the discriminant determines the number and nature of the solutions:
  • If it's positive, there are two distinct real solutions.
  • If it's zero, there is one real solution.
  • If it's negative, there are no real solutions, but two complex ones.

Using the quadratic formula, you can find the roots of equations like \(2x^2 - 3x + 1 = 0\), which simplifies the problem of finding angle values such as \(\theta\) achieved in trigonometric equations.