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To find where a function has a horizontal tangent, it’s essential to understand the role of the derivative. The derivative of a function provides us with the slope of the tangent line at any given point on the function’s graph. When the derivative equals zero, this indicates that the tangent is horizontal at that point. This is because a zero slope means the line is flat, neither rising nor falling.

For the function given in the exercise, \( y = x + 2 \cos x \), we calculate the derivative using basic rules of differentiation. The derivative of \( x \) is simply 1, and for \( 2 \cos x \), we apply the derivative rule for cosine, which is \(-\sin x\). Thus, the derivative of \( y \) becomes \( \frac{dy}{dx} = 1 - 2 \sin x \).

Identifying where this derivative equals zero leads us to the points with horizontal tangents. This step is crucial in any analysis involving tangents, as it transforms a visual question into an algebraic one, allowing us to pinpoint exact values of \( x \) for further consideration.

Trigonometric functions like \( \cos x \) and \( \sin x \) appear regularly in calculus problems, often due to their periodic nature and interesting properties. In this context, \( \cos x \) and \( \sin x \) contribute to computing areas of interest, such as points of horizontal tangency.

For the function \( y = x + 2 \cos x \), \( \sin x \) emerges when differentiating \( \cos x \). The specific relation needed here is where the sine function itself hits the value \( \frac{1}{2} \). This occurs because we set \( 1 - 2 \sin x = 0 \), leading to \( \sin x = \frac{1}{2} \).

\( \sin x \) is \( \frac{1}{2} \) at specific angles in its cycle: \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \), within the interval \(0 \leq x \leq 2\pi\). These values stem from the known unit circle values and symmetry of the sine function, which repeat every \( 2\pi \). Understanding these properties of trig functions is key in solving problems like these, where algebra meets geometry.

Interval analysis helps us determine the region where a solution is valid. In problems dealing with horizontal tangents, it's important to focus on where the derivative solves the equation within a specified range. Here, the interval is \( 0 \leq x \leq 2\pi \), covering one full cycle of typical trigonometric functions.

When we solve \( 1 - 2 \sin x = 0 \) for \( \sin x = \frac{1}{2} \), we check for solutions within the given interval. We find values \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \), which lie within our range of interest. These values represent the locations where our function has horizontal tangents.

It's crucial to always verify that any obtained solutions fit within the predefined interval. This ensures that the results are consistent with any constraints periodical or defined by the problem. Interval analysis provides clarity especially when multiple solutions might exist outside of your range. It can be the difference between a correct and incorrect understanding of function behavior.