Question

The general solution of the equation \(cotθ\cdot cot2θ = 1\) is...

• A. \(θ = nπ\pm \frac{π}{6},n\in Z\)
• B. \(θ = nπ\pm \frac{π}{3},n\in Z\)
• C. \(θ = nπ\pm \frac{π}{4},n\in Z\)
• D. \(θ = nπ\pm \frac{π}{8},n\in Z\)

Hint:

Convert to cos3 theta = 0 and then remove the values where cot 2 theta is undefined.

Answer 1

The Correct Option is A

Step 1: Rewrite the equation
\(\cot\theta\cot2\theta = 1\) gives \(\cos\theta\cos2\theta = \sin\theta\sin2\theta\), so \(\cos\theta\cos2\theta - \sin\theta\sin2\theta = 0\), that is \(\cos3\theta = 0\).

Step 2: Solve cos 3 theta = 0
\(3\theta = (2m+1)\frac{\pi}{2}\), so \(\theta = (2m+1)\frac{\pi}{6}\), with values \(\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{7\pi}{6},\ldots\).

Step 3: Remove invalid values
At \(\theta = \frac{\pi}{2}+k\pi\), \(\cot2\theta\) is undefined because \(\sin2\theta = 0\). These must be discarded.

Step 4: Combine the rest
The remaining values are \(\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6},\ldots\), which are \(n\pi\pm\frac{\pi}{6}\). This is option (A). Options (B), (C) and (D) give angles such as \(\frac{\pi}{3}\), where \(\cot\theta\cot2\theta = \frac{1}{\sqrt3}\cdot\left(-\frac{1}{\sqrt3}\right) \ne 1\).

Final Answer:
The general solution is theta = n pi plus or minus pi/6. \[ \boxed{\text{(A)}\ \theta = n\pi\pm\frac{\pi}{6}} \]