Problem 11 - Algebra and Trigonometry

Question

\(3-16 \cdot\) Solve the given equation. $$ 2 \sin ^{2} \theta-\cos \theta=1 $$

Step-by-Step Solution

Answer

The solutions are \( \theta = \frac{\pi}{3}, \frac{5\pi}{3}, \pi \).

Step 1: Use Trigonometric Identity

Begin by using the Pythagorean identity to express everything in terms of one trigonometric function. The identity we use is: \( \sin^2 \theta = 1 - \cos^2 \theta \). Substitute this into the equation:\[ 2(1 - \cos^2 \theta) - \cos \theta = 1 \].

Step 2: Simplify the Equation

Expand and simplify the equation obtained by substituting the trigonometric identity:\[ 2 - 2\cos^2 \theta - \cos \theta = 1 \].Next, move all terms to one side of the equation:\[ 2\cos^2 \theta + \cos \theta - 1 = 0 \].

Step 3: Solve the Quadratic Equation

The equation \( 2\cos^2 \theta + \cos \theta - 1 = 0 \) is a quadratic equation in terms of \( \cos \theta \). Solve it using the quadratic formula:\[ \cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \],where \( a = 2 \), \( b = 1 \), \( c = -1 \).

Step 4: Calculate the Roots

Substitute \( a = 2 \), \( b = 1 \), and \( c = -1 \) into the quadratic formula:\[ \cos \theta = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]\[ \cos \theta = \frac{-1 \pm \sqrt{1 + 8}}{4} \]\[ \cos \theta = \frac{-1 \pm 3}{4} \].

Step 5: Determine Possible Solutions for Cosine

Simplify to get the possible values of \( \cos \theta \):\[ \cos \theta = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad \cos \theta = \frac{-4}{4} = -1 \].

Step 6: Find \( \theta \) from Cosine Values

For \( \cos \theta = \frac{1}{2} \), \( \theta = \frac{\pi}{3} , \frac{5\pi}{3} \). For \( \cos \theta = -1 \), \( \theta = \pi \). These angles satisfy the cosine values within the interval \( [0, 2\pi) \).

Step 7: Validate Solutions in Original Equation

Validate that all these angles satisfy the original equation:- For \( \theta = \frac{\pi}{3} \), \(2 \sin^2 \frac{\pi}{3} - \cos \frac{\pi}{3} = \frac{3}{2} - \frac{1}{2} = 1 \), so it satisfies the equation.- For \( \theta = \frac{5\pi}{3} \), it similarly satisfies the equation.- For \( \theta = \pi \), \(2 \sin^2 \pi - \cos \pi = 0 + 1 = 1 \), it also satisfies the equation.

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools in solving equations involving trigonometric functions. They allow us to express one function in terms of another, thereby simplifying the problem. In this exercise, we apply the Pythagorean identity:- \( \sin^2 \theta = 1 - \cos^2 \theta \)This identity is key, because it enables the conversion of the original equation \( 2 \sin^2 \theta - \cos \theta = 1 \) into another form entirely in terms of cosine. By replacing \( \sin^2 \theta \) with \( 1 - \cos^2 \theta \), the equation becomes:- \[ 2(1 - \cos^2 \theta) - \cos \theta = 1 \] Such transformations using identities are often the first step, setting the stage for further simplifications or substitutions. Understanding and recognizing different trigonometric identities is crucial for efficiently solving these types of equations.

Quadratic Equations in Trigonometry

Once an equation is transformed entirely in terms of a single trigonometric function, it often resembles a quadratic equation. This is the case with the equation derived from the original problem: - \[ 2\cos^2 \theta + \cos \theta - 1 = 0 \] A quadratic equation in the standard form has the expression \( ax^2 + bx + c = 0 \). Here, the trigonometric function \( \cos \theta \) acts as the variable. To find the roots of such equations, we typically use the quadratic formula: - \( \cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) Substituting our specific values \( a = 2, b = 1, \) and \( c = -1 \), allows us to solve for the cosine values:- \[ \cos \theta = \frac{-1 \pm \sqrt{1 + 8}}{4} \] - \( \cos \theta = \frac{-1 \pm 3}{4} \)The solutions here are \( \cos \theta = \frac{1}{2} \) and \( \cos \theta = -1 \), which lead us to potential angles for \( \theta \). Correctly identifying and solving quadratic equations in this context is essential for finding the correct angles.

Solution Verification

Once potential solutions are found, it is important to verify them within the original equation. Verification ensures that no mistakes were made and confirms the legitimacy of the solutions.Given the solutions \( \theta = \frac{\pi}{3}, \frac{5\pi}{3}, \text{ and } \pi \), each must be plugged back into the original equation:- For \( \theta = \frac{\pi}{3} \), substitute and simplify: \[ 2\sin^2 \frac{\pi}{3} - \cos \frac{\pi}{3} = \frac{3}{2} - \frac{1}{2} = 1 \] This confirms that it satisfies the equation.- Similarly, for \( \theta = \frac{5\pi}{3} \): \[ 2\sin^2 \frac{5\pi}{3} - \cos \frac{5\pi}{3} = \frac{3}{2} - \frac{1}{2} = 1 \] This solution also checks out.- For \( \theta = \pi \): \[ 2\sin^2 \pi - \cos \pi = 0 + 1 = 1 \] Again, the solution holds true.Verifying each solution like this precludes any oversight and builds confidence in the accuracy of the results.