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When dealing with equations like \[(\tan ^{2} \theta-4)(2 \cos \theta+1)=0\], it's important to recognize that the equation is in a "product equals zero" form. This means that for the entire product to be zero, at least one of the factors must be zero.

To solve such equations, we break them down into simpler parts:

• Start by identifying the individual factors: \(\tan^2 \theta - 4\) and \(2 \cos \theta + 1\).
• Set each factor equal to zero separately to find possible solutions.
  • For the first factor: \(\tan^2 \theta - 4 = 0\).
  • For the second factor: \(2 \cos \theta + 1 = 0\).

Once each equation is solved separately, the comprehensive solution set of the original equation is the combination of the solutions from both equations.

Trigonometric functions are essential tools in mathematics, particularly when dealing with angles and triangles. In the context of this exercise, the trigonometric functions involved are the tangent and cosine functions.

The tangent function, \(\tan \theta\), relates the sine and cosine functions as \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). The square of this function, \(\tan^2 \theta = 4\), leads to solutions by taking square roots, which introduces the possibility of both positive and negative results.

Similarly, the cosine function, \(\cos \theta\), is a basic trigonometric function representing the ratio of the adjacent side to the hypotenuse in a right-angled triangle. Solving \(2 \cos \theta + 1 = 0\) requires manipulating the equation to express \(\cos \theta\) in terms of standard angles. Understanding these functions deeply helps in solving and interpreting the behavior of trigonometric equations.

The relationship between tangent and cosine functions is pivotal for solving the given equation. In this specific exercise, we explore the behavior of both functions under specific conditions.

When \(\tan^2 \theta = 4\), we know that \(\tan \theta = \pm 2\), leading directly to \(\theta = \arctan(2) + k\pi\) or \(\theta = \arctan(-2) + k\pi\) for integer \(k\). This takes into account the tangent function's periodicity, repeating every \(\pi\) units.

On the other hand, solving \(\cos \theta = -\frac{1}{2}\) involves identifying angles where the cosine function yields \(-\frac{1}{2}\). This occurs at angles like \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\), though the function repeats every \(2\pi\). Both solutions reinforce the properties of tangent and cosine functions, illustrating how periodic and specific they are with respect to angle measurements.