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: Use tangent form
With $t = \tan\theta$, $\cot2\theta = \frac{1-t^2}{2t}$, so the equation is $\frac1t\cdot\frac{1-t^2}{2t} = 1$.

Step 2: Solve for t
$1-t^2 = 2t^2$ gives $t^2 = \frac13$, so $\tan\theta = \pm\frac{1}{\sqrt3}$.

Step 3: General solution
$\tan^2\theta = \tan^2\frac{\pi}{6}$ gives $\theta = n\pi\pm\frac{\pi}{6}$, option (A).

Final Answer:
Option A. \[ \boxed{\text{(A)}\ \theta = n\pi\pm\frac{\pi}{6}} \]