Question

The general solution of the equation \( \sin^2 \theta + 3 \cos^2 \theta = 5 \sin \theta \) is:

• A. \( n\pi + \frac{\pi}{3}, n \in \mathbb{Z} \)
• B. \( n\pi + (-1)^n \frac{\pi}{6}, n \in \mathbb{Z} \)
• C. \( n\pi \pm \frac{\pi}{6}, n \in \mathbb{Z} \)
• D. \( n\pi + (-1)^n \frac{\pi}{3}, n \in \mathbb{Z} \)

Hint:

When solving trigonometric equations, use known values of sine and cosine, and don't forget to apply the periodicity of the trigonometric functions to obtain the general solution.

Answer 1

The Correct Option is B

Step 1: Rewrite the equation using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \). \[ \sin^2 \theta + 3(1 - \sin^2 \theta) = 5 \sin \theta \] \[ \sin^2 \theta + 3 - 3 \sin^2 \theta = 5 \sin \theta \] \[ 3 - 2 \sin^2 \theta = 5 \sin \theta \] \[ 2 \sin^2 \theta + 5 \sin \theta - 3 = 0 \] Step 2: Solve the quadratic equation \( 2 \sin^2 \theta + 5 \sin \theta - 3 = 0 \). We will solve it using the quadratic formula. The equation is in the form \( a \sin^2 \theta + b \sin \theta + c = 0 \), where: \[ a = 2, \, b = 5, \, c = -3 \] Using the quadratic formula: \[ \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] \[ \sin \theta = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)} \] \[ \sin \theta = \frac{-5 \pm \sqrt{25 + 24}}{4} \] \[ \sin \theta = \frac{-5 \pm \sqrt{49}}{4} \] \[ \sin \theta = \frac{-5 \pm 7}{4} \] Thus, the two possible values for \( \sin \theta \) are: \[ \sin \theta = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}, \quad \sin \theta = \frac{-5 - 7}{4} = \frac{-12}{4} = -3 \] Since \( \sin \theta = -3 \) is not a valid solution (because \( \sin \theta \) must lie between -1 and 1), we have: \[ \sin \theta = \frac{1}{2} \] Step 3: Determine the values of \( \theta \) that satisfy \( \sin \theta = \frac{1}{2} \). The general solutions for \( \sin \theta = \frac{1}{2} \) are: \[ \theta = \frac{\pi}{6} + 2n\pi \quad {or} \quad \theta = \pi - \frac{\pi}{6} + 2n\pi \] Simplifying: \[ \theta = \frac{\pi}{6} + 2n\pi \quad {or} \quad \theta = \frac{5\pi}{6} + 2n\pi \] This can be written as: \[ \theta = n\pi + (-1)^n \frac{\pi}{6}, \quad n \in \mathbb{Z} \] Step 4: Conclusion Thus, the general solution is: \[ \boxed{n\pi + (-1)^n \frac{\pi}{6}, \, n \in \mathbb{Z}} \]