Question

The general solution of the equation \(4\,\sin x\cos x+2\,\sin x+2\,\cos x+1=0\), where \(n\in Z\) is:

• A. \(x=2n\pi \pm \frac{2\pi }{3}\) or \(n\pi +(-1)^{n}\frac{7\pi }{6}\)
• B. \(x=2n\pi \pm \frac{\pi }{3}\) or \(n\pi -(-1)^{n}\frac{5\pi }{6}\)
• C. \(x=2n\pi +\frac{\pi }{3}\) or \(n\pi +(-1)^{n}\frac{7\pi }{6}\)
• D. \(x=2n\pi -\frac{\pi }{3}\) or \(n\pi +(-1)^{n}\frac{7\pi }{6}\)

Hint:

The left side factors as (2 sin x + 1)(2 cos x + 1), so solve each factor separately.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Factor the expression by grouping.

Step 2: Factorisation
\(4\sin x\cos x+2\sin x+2\cos x+1=(2\sin x+1)(2\cos x+1)=0\)
So \(\sin x=-\frac{1}{2}\) or \(\cos x=-\frac{1}{2}\).

Step 3: Solve each
If \(\sin x=-\frac{1}{2}\), then \(\sin x=\sin\frac{7\pi}{6}\), so \(x=n\pi+(-1)^n\frac{7\pi}{6}\).
If \(\cos x=-\frac{1}{2}\), then \(\cos x=\cos\frac{2\pi}{3}\), so \(x=2n\pi\pm\frac{2\pi}{3}\).

Step 4: Compare the options
Both solution sets together match the first option.

Final Answer:
The solution is x = 2nπ ± 2π/3 or x = nπ + (-1)^n 7π/6. \[ \boxed{\text{(A) }x=2n\pi\pm\dfrac{2\pi}{3}\text{ or }n\pi+(-1)^{n}\dfrac{7\pi}{6}} \]