The solution of the equation \(\tan 3 \theta+\cot \theta=0\) is (a) \(\\{(2 k+1)(\pi / 2), k \in z\\}\) (b) \(\\{k \pi, k \in z\\}\) (c) \(\\{(2 k+1)(\pi / 4), k \in z\\}\) (d) \(\\{(2 k+1)(\pi / 6), k \in z\\}\)

Recall that \(\tan(x)=\frac{\sin(x)}{\cos(x)}\) and \(\cot(x)=\frac{\cos(x)}{\sin(x)}\). Our equation becomes: \[ \frac{\sin(3\theta)}{\cos(3\theta)} + \frac{\cos(\theta)}{\sin(\theta)} = 0 \]

To add the fractions on the left side, we need to equate the denominators. Multiply the first fraction by \(\sin(\theta)\cos(\theta)\) and the second fraction by \(\sin(3\theta)\cos(3\theta)\). The equation becomes: \[ \sin(\theta)\cos(\theta)\frac{\sin(3\theta)}{\cos(3\theta)} + \sin(3\theta)\cos(3\theta)\frac{\cos(\theta)}{\sin(\theta)} = 0 \] Simplify further to obtain: \[ \sin(3\theta)\sin(\theta)\cos(\theta) + \sin(3\theta)\cos(3\theta)\cos(\theta) = 0 \]

We notice that there's a common factor of \(\sin(3\theta)\cos(\theta)\) in both terms. Factor this out to get: \[ \sin(3\theta)\cos(\theta)(\sin(\theta) + \cos(3\theta)) = 0 \]

Now, we must solve for \(\theta\) when any of the factors is equal to 0. 1. \(\sin(3\theta) = 0\): This is true when \(3\theta = k\pi\), where k is an integer. Therefore, \(\theta = \frac{k\pi}{3}\). 2. \(\cos(\theta) = 0\): This is true when \(\theta = (2k+1)\frac{\pi}{2}\), where k is an integer. 3. \(\sin(\theta) + \cos(3\theta) = 0\): We don't have a direct way to solve this analytically, but we can notice that the given answer choices do not include any combinations of \(\theta\) resulting from this factor. Now, combining the solutions from the first and second factors, we write the general solution for \(\theta\) as \(\theta = \frac{k\pi}{3}(2k+1)\frac{\pi}{2}\), where k is an integer. This general solution corresponds to option (a): \(\{(2k+1)(\pi/2), k \in \mathbb{Z}\}\).