Solve the trigonometric equation \[\sin 2\theta + \cos 2\theta = 0\]
A). \[\theta = 2n\pi - \dfrac{\pi }{3},n \in Z\]
B). \[\theta = \dfrac{{2n\pi }}{3} - \dfrac{\pi }{4},n \in Z\]
C). \[\theta = 2n\pi - \dfrac{\pi }{2},n \in Z\]
D). \[\theta = \dfrac{{2n\pi }}{3} - \dfrac{\pi }{6},n \in Z\]

Hint: In the given question, we have been given an expression which has two trigonometric terms with the same argument (or angle). We have been given their sum. We have to find the measure of the argument. We are going to solve it by dividing both sides by the square root of the sum of squares of coefficients of the arguments. Then we are going to convert the trigonometric terms into known values from the standard value table. Then, we will apply the appropriate formula to condense the trigonometric terms into some basic trigonometric identities and solve for it.

Complete step-by-step solution:
The given expression is \[\sin 2\theta + \cos 2\theta = 0\].
\[ \Rightarrow \cos 2\theta = - \sin 2\theta \]
We know,
\[ - \sin 2\theta = \cos \left( {\dfrac{\pi }{2} + 2\theta } \right)\]
So, we have,
\[\theta = 2n\pi - \left( {\dfrac{\pi }{2} + 2\theta } \right),n \in Z\]
Now, we have,
\[\theta = 2n\pi - \dfrac{\pi }{2} - 2\theta \]
Taking \[2\theta \] to the other side,
\[3\theta = 2n\pi - \dfrac{\pi }{2}\]
Taking the \[3\] multiplied with \[\theta \] to the other side, we have,
\[\theta = \dfrac{{2n\pi }}{3} - \dfrac{\pi }{6}\]
Hence, the correct option is D.

Note: In the given question, we were given a trigonometric expression in which there were two terms whose angle was the same, and they were being added. We were given their sum and had to find the angle. We did it by converting the given expression into the result of a standard trigonometric formula, then we condensed it to the formula by substituting the values and found the answer to the question. So, it is very important that we have enough practice of such questions so that we know how to convert an expression into a formula and use it to find the answer to our question.