Question

The general solution of $ 2 \cos 4x + \sin^2 2x = 0 \text{ is}

• A. \( x = \frac{\pi}{4} \pm \sin^{-1} \left( \frac{1}{3} \right) \)
• B. \( x = \frac{\pi}{4} + (-1)^n \sin^{-1} \left( \pm \frac{\sqrt{2}}{3} \right) \)
• C. \( x = \frac{\pi}{2} \pm \cos^{-1} \left( \frac{1}{5} \right) \)
• D. \( x = \frac{\pi}{4} + (-1)^n \cos^{-1} \left( \frac{1}{5} \right) \)

Hint:

When solving trigonometric equations, always look for possible identities or simplifications to reduce the complexity of the equation.

Answer 1

The Correct Option is B

The given equation is: \[ 2 \cos 4x + \sin^2 2x = 0 \] We can rewrite this equation by using a trigonometric identity: \[ \sin^2 2x = 1 - \cos^2 2x \] Substitute this into the equation: \[ 2 \cos 4x + 1 - \cos^2 2x = 0 \] Now, simplify the equation and solve for \( x \). After simplifying, we get the general solution: \[ x = \frac{\pi}{4} + (-1)^n \sin^{-1} \left( \pm \frac{\sqrt{2}}{3} \right) \] Thus, the correct solution is option (B).