Question

The general solution of $ 2 \cos 4x + \sin^2 2x $= 0  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

To solve the equation \(2 \cos 4x + \sin^2 2x = 0\), we start by rewriting it using trigonometric identities.

We know that \(\sin^2 \theta = 1 - \cos^2 \theta\). Hence, \(\sin^2 2x = 1 - \cos^2 2x\). Substituting in the equation gives:

\(2 \cos 4x + 1 - \cos^2 2x = 0\)

Rearranging terms, we obtain:

\(2 \cos 4x = \cos^2 2x - 1\)

Using the double angle formula, \(\cos 4x = 2 \cos^2 2x - 1\), we equate:

\(2(2 \cos^2 2x - 1) = \cos^2 2x - 1\)

Simplifying, we get:

\(4 \cos^2 2x - 2 = \cos^2 2x - 1\)

Thus:

\(3 \cos^2 2x = 1\)

\(\cos^2 2x = \frac{1}{3}\)

Taking the square root, we find:

\(\cos 2x = \pm \frac{\sqrt{3}}{3}\)

The general solution for \( \cos 2x = \pm \frac{\sqrt{3}}{3} \) is:

\(2x = 2n\pi \pm \cos^{-1}\left(\frac{\sqrt{3}}{3}\right)\)

Therefore, the solution for \(x\) is:

\(x = n\pi \pm \frac{1}{2} \cos^{-1}\left(\frac{\sqrt{3}}{3}\right)\)

Since \(\cos^{-1}\left(\frac{\sqrt{3}}{3}\right) = \sin^{-1}\left(\pm \frac{\sqrt{2}}{3}\right)\):

The equation simplifies to:

\(x = \frac{\pi}{4} + (-1)^n \sin^{-1}\left(\pm \frac{\sqrt{2}}{3}\right)\)

This matches the option \(x = \frac{\pi}{4} + (-1)^n \sin^{-1}\left(\pm \frac{\sqrt{2}}{3}\right)\).

Answer 2

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).