Question

The sum of all values of \( \theta \in [0, 2\pi] \) satisfying \( 2\sin^2\theta = \cos 2\theta \) and \( 2\cos^2\theta = 3\sin\theta \) is:

• A. \( \frac{\pi}{2} \)
• B. \( 4\pi \)
• C. \( \pi \)
• D. \( \frac{5\pi}{6} \)

Hint:

To solve trigonometric equations involving identities like \( \cos 2\theta \) and \( \sin^2\theta \), - Use standard trigonometric identities and algebraic manipulations to simplify the equations. - Solving step by step and checking for all possible values of \( \theta \) in the given range will help in obtaining the correct sum of solutions.

Answer 1

The Correct Option is C

We are tasked with solving for \( \theta \) in the interval \([0, 2\pi]\) using two given equations:

1. \( 2\sin^2\theta = \cos 2\theta \)
2. \( 2\cos^2\theta = 3\sin\theta \)

Let's solve each equation independently:

Equation 1: \( 2\sin^2\theta = \cos 2\theta \)

Using the double angle identity \(\cos 2\theta = 1-2\sin^2\theta\), we substitute it into the equation:

\( 2\sin^2\theta = 1 - 2\sin^2\theta \)

Adding \( 2\sin^2\theta \) to both sides gives:

\( 4\sin^2\theta = 1 \)

Dividing by 4 yields:

\( \sin^2\theta = \frac{1}{4} \)

Taking the square root, we get:

\( \sin\theta = \frac{1}{2} \) or \( \sin\theta = -\frac{1}{2} \)

The solutions in the specified interval are \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\) for \( \sin\theta = \frac{1}{2} \) and \(\theta = \frac{7\pi}{6}, \frac{11\pi}{6}\) for \(\sin\theta = -\frac{1}{2}\).

Equation 2: \( 2\cos^2\theta = 3\sin\theta \)

Substituting \(\cos^2\theta = 1-\sin^2\theta\) results in:

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

\( 2 - 2\sin^2\theta = 3\sin\theta \)

Rearranging the terms to form a quadratic equation in terms of \(\sin\theta\):

\( 2\sin^2\theta + 3\sin\theta - 2 = 0 \)

Let \( x = \sin\theta \). The equation becomes:

\( 2x^2 + 3x - 2 = 0 \)

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \):

\( x = \frac{-3 \pm \sqrt{3^2 - 4\cdot2\cdot(-2)}}{2\cdot2} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4} \)

This gives two possible values for \( x \): \( x = \frac{1}{2} \) or \( x = -2 \). Since the range of \(\sin\theta\) is \([-1, 1]\), \( \sin\theta = -2 \) is not a valid solution. Therefore, we consider only \(\sin\theta = \frac{1}{2}\).

Intersection of Solutions:

The common valid solutions for \(\sin\theta = \frac{1}{2}\) derived from both equations are \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\).

Sum of all solutions:

\(\frac{\pi}{6} + \frac{5\pi}{6} = \frac{6\pi}{6} = \pi\)

The sum of all valid solutions is \(\pi\).