Question

Find the general solution of \( \sin 2x + \cos x = 0 \).

• A. \( (n \pm 1)\frac{\pi}{2} \)
• B. \( n\pi \pm \frac{\pi}{2} \)
• C. \( n\pi + (-1)^n \frac{7\pi}{6} \) or \( (2n + 1)\frac{\pi}{2} \)
• D. None of the above

Hint:

When solving trigonometric equations like \( \sin 2x + \cos x = 0 \), factorize using identities such as \( \sin 2x = 2 \sin x \cos x \). Then solve each factor separately to get the general solutions.

Answer 1

The Correct Option is C

Step 1: Use the identity for \( \sin 2x \)
\[ \sin 2x + \cos x = 0 \Rightarrow 2 \sin x \cos x + \cos x = 0 \] Step 2: Take common factor
\[ \cos x (2 \sin x + 1) = 0 \] Step 3: Solve each factor separately
Case 1: \( \cos x = 0 \)
\[ x = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \Rightarrow x = (2n + 1)\frac{\pi}{2} \] Case 2: \( 2 \sin x + 1 = 0 \Rightarrow \sin x = -\frac{1}{2} \)
\[ x = \sin^{-1} \left(-\frac{1}{2} \right) = -\frac{\pi}{6} \] Step 4: General solution for \( \sin x = -\frac{1}{2} \)
\[ x = n\pi + (-1)^n \left(-\frac{\pi}{6}\right) = n\pi + (-1)^{n+1} \frac{\pi}{6} \Rightarrow x = n\pi + (-1)^n \frac{7\pi}{6} \quad (\text{since } -\frac{\pi}{6} \equiv \frac{11\pi}{6}, \text{ same modulo } 2\pi) \] Thus, the complete general solution is:
\[ x = (2n + 1)\frac{\pi}{2} \quad \text{or} \quad x = n\pi + (-1)^n \frac{7\pi}{6} \]