Question

The number of solutions of \[ \sin 2x + \cos 4x = 2 \quad \text{in the interval } [-\pi, \pi] \text{ is:} \]

• A. \( 3 \)
• B. \( 2 \)
• C. \( 0 \)
• D. \( 1 \)

Hint:

For equations involving sums of bounded trigonometric functions, check the conditions under which the sum reaches maximum value.

Answer 1

The Correct Option is C

Step 1: Maximum value of \( \sin 2x \) is 1, and maximum value of \( \cos 4x \) is also 1. 
Therefore, 
\[ \sin 2x + \cos 4x \leq 1 + 1 = 2 \] Step 2: Equality holds only if \( \sin 2x = 1 \) and \( \cos 4x = 1 \) 
\[ \sin 2x = 1 \Rightarrow 2x = \dfrac{\pi}{2} + 2n\pi \Rightarrow x = \dfrac{\pi}{4} + n\pi \] \[ \cos 4x = 1 \Rightarrow 4x = 2m\pi \Rightarrow x = \dfrac{m\pi}{2} \] Equating: 
\[ \dfrac{\pi}{4} + n\pi = \dfrac{m\pi}{2} \Rightarrow \text{Check for integer solutions in } [-\pi, \pi] \] This equation has no solution for integer \( n, m \) such that \( x \in [-\pi, \pi] \) 
Hence, no value of \( x \) in \( [-\pi, \pi] \) satisfies both conditions simultaneously.