Question

If \[ 7 \cos^2 x + 3 \sin^2 x = 6, \] then the value of \( \cos 2x \) is equal to

• A. \( \frac{1}{2} \)
• B. \( \frac{3}{2} \)
• C. \( \frac{5}{2} \)
• D. \( 1 \)
• E. \( 2 \)

Hint:

Express squares of trigonometric functions in terms of double angles to simplify equations.

Answer 1

The Correct Option is A

We use the identity: \[ \cos^2 x = \frac{1 + \cos 2x}{2}, \quad \sin^2 x = \frac{1 - \cos 2x}{2} \] Rewriting the given equation: \[ 7 \cos^2 x + 3 \sin^2 x = 6 \] Substituting the identities: \[ 7 \times \frac{1 + \cos 2x}{2} + 3 \times \frac{1 - \cos 2x}{2} = 6 \] Expanding: \[ \frac{7 + 7 \cos 2x + 3 - 3 \cos 2x}{2} = 6 \] \[ \frac{10 + 4 \cos 2x}{2} = 6 \] \[ 10 + 4 \cos 2x = 12 \] \[ 4 \cos 2x = 2 \] \[ \cos 2x = \frac{1}{2} \] Final Answer: \[ \boxed{\frac{1}{2}} \]