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 given two equations to solve for \( \theta \) in the interval \([0, 2\pi]\):

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

Let's analyze each equation:

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

We know \(\cos 2\theta = 1-2\sin^2\theta\) from the double angle identity. Substitute it into the equation:

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

Add \( 2\sin^2\theta \) to both sides:

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

Divide by 4:

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

Taking the square root yields:

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

Values 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 \(\cos^2\theta = 1-\sin^2\theta\) gives:

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

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

Rearrange:

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

Let \( x = \sin\theta \), then:

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

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

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

Thus, \( x = \frac{1}{2} \) or \( x = -2 \). Since \(\sin\theta\) cannot be -2, only \(\sin\theta = \frac{1}{2}\) is valid.

Intersection of Solutions:

From both equations, valid solutions for \(\sin\theta = \frac{1}{2}\) are \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\).

Sum of all solutions:

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

Thus, the sum of all solutions is \(\pi\).

Answer 2

Step 1: Simplify the first equation.
Using \( \cos 2\theta = 1 - 2\sin^2\theta \), the equation \[ 2\sin^2\theta = \cos 2\theta \] becomes \[ 2\sin^2\theta = 1 - 2\sin^2\theta \;\Longrightarrow\; 4\sin^2\theta = 1 \;\Longrightarrow\; \sin\theta = \pm \tfrac{1}{2}. \]

Step 2: Simplify the second equation.
From \( \cos^2\theta = 1 - \sin^2\theta \), the equation \[ 2\cos^2\theta = 3\sin\theta \] gives \[ 2(1 - \sin^2\theta) = 3\sin\theta \;\Longrightarrow\; 2 - 2\sin^2\theta = 3\sin\theta. \] Let \( s = \sin\theta \). Then \[ 2s^2 + 3s - 2 = 0 \;\Longrightarrow\; s = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}. \] Hence \( s = \tfrac{1}{2} \) or \( s = -2 \). Since \( \sin\theta \in [-1,1] \), the only admissible solution is \[ \sin\theta = \tfrac{1}{2}. \]

Step 3: Common solutions to both equations.
From Step 1, \( \sin\theta = \pm \tfrac{1}{2} \), and from Step 2, \( \sin\theta = \tfrac{1}{2} \).
Therefore, the common solutions satisfy \( \sin\theta = \tfrac{1}{2} \) in \( [0, 2\pi] \): \[ \theta = \frac{\pi}{6},\; \frac{5\pi}{6}. \]

Step 4: Sum of all solutions.
\[ \frac{\pi}{6} + \frac{5\pi}{6} = \pi. \]

Final Answer

\[ \boxed{\pi} \]