Problem 43 - Precalculus Mathematics for Calculus

Question

Solve the given equation. $$3 \sin ^{2} \theta-7 \sin \theta+2=0$$

Step-by-Step Solution

Answer

\( \theta = \arcsin \frac{1}{3} + 2k\pi \) or \( \theta = \pi - \arcsin \frac{1}{3} + 2k\pi \).

Step 1: Identify substitution

To simplify the equation, let's set a substitution. Let \( x = \sin \theta \). The equation becomes: \[ 3x^2 - 7x + 2 = 0 \]

Step 2: Apply the quadratic formula

The equation \( 3x^2 - 7x + 2 = 0 \) is a quadratic equation in the form \( ax^2 + bx + c = 0 \). Use the quadratic formula to solve for \( x \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a = 3 \), \( b = -7 \), and \( c = 2 \). Plug these into the formula.

Step 3: Calculate the discriminant

First, calculate the discriminant \( b^2 - 4ac \):\[ (-7)^2 - 4 \times 3 \times 2 = 49 - 24 = 25 \].Since the discriminant is a perfect square, there will be two real roots.

Step 4: Solve for roots

Substitute the values back into the quadratic formula:\[ x = \frac{-(-7) \pm \sqrt{25}}{6} = \frac{7 \pm 5}{6} \].This results in two solutions: \( x = \frac{12}{6} = 2 \) and \( x = \frac{2}{6} = \frac{1}{3} \).

Step 5: Relate roots back to \( \sin \theta \)

Since \( x = \sin \theta \), we have \( \sin \theta = 2 \) and \( \sin \theta = \frac{1}{3} \). The value \( \sin \theta = 2 \) is not possible since sine values range from \(-1\) to \(1\). Thus, only \( \sin \theta = \frac{1}{3} \) is valid.

Step 6: Solve for \( \theta \)

Find \( \theta \) such that \( \sin \theta = \frac{1}{3} \). \( \theta = \arcsin \frac{1}{3} \)This equation has solutions of the form \( \theta = \arcsin \frac{1}{3} + 2k\pi \) and \( \theta = \pi - \arcsin \frac{1}{3} + 2k\pi \) for integer \( k \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a handy technique to transform complex trigonometric equations into simpler ones, often resembling algebraic equations. In this particular problem, the original equation was presented in terms of \( \sin \theta \). By letting \( x = \sin \theta \), the equation

• Instead of handling a trigonometric form directly, this transformation allows us to work with a quadratic expression \( 3x^2 - 7x + 2 = 0 \),
• reduces the initial trigonometric equation to something more familiar and easier to solve.

After solving the quadratic expression, the roots found need to be related back to the trigonometric terms using the initial substitution.

Quadratic Formula

The quadratic formula is a vital tool in solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It provides a straightforward method to find the values of \( x \) for which the equation holds true. The general formula is:

• \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
• This formula encompasses all possible solutions by utilizing the square root, represented by \( \pm \).

For the exercise, substituting \( a = 3 \), \( b = -7 \), and \( c = 2 \) into the quadratic formula helps calculate the values of \( x \) that are potential solutions to the trigonometric equation once back substitution is employed.

Solving Trigonometric Equations

Solving trigonometric equations involves several steps to ensure all potential solutions are identified. In this exercise:

• Start by transforming the trigonometric equation using substitution, making it a quadratic equation in terms of \( x \), where \( x = \sin \theta \).
• Find the solution to this quadratic equation using the quadratic formula. In this case, two potential solutions were identified: \( x = 2 \) and \( x = \frac{1}{3} \).
• Check each potential solution to ensure they fall within the sine function's range, which is from \(-1\) to \(1\).

Since \( \sin \theta = 2 \) is outside this range, only \( \sin \theta = \frac{1}{3} \) is a valid root. The solution proceeds by finding the angle \( \theta \) such that \( \sin \theta = \frac{1}{3} \).

Discriminant in Quadratics

The discriminant is a critical component of the quadratic formula's expression under the square root, specifically \( b^2 - 4ac \). It indicates the nature and number of roots of a quadratic equation.

• A positive discriminant, such as \( 25 \) in this problem, suggests two distinct real solutions, which means there are potentially two valid angles for \( \theta \) to solve.
• If the discriminant were zero, it would imply a single repeated real solution.
• A negative discriminant indicates no real solutions, meaning we deal with complex numbers instead.

Understanding the discriminant helps to manage expectations about the number and type of solutions before fully solving the quadratic equation. In trigonometric equations, this can be crucial for identifying valid angle solutions. By calculating \( b^2 - 4ac = 25 \), we know the equation has two real roots, assisting us in solving for \( \sin \theta \). Since one root did not satisfy the sine function's limits, the focus remained on the valid root, linking to \( \theta = \arcsin \frac{1}{3} \).