91Ó°ÊÓ

Differentiation is a fundamental concept in calculus. It is the process of finding the derivative of a function, which measures how the function's value changes as its input changes. The derivative tells us the rate of change or the slope of the function at any given point.
In the exercise provided, the function in question is \(f(x) = x + 2 \sin x\). To find its derivative, we use the basic rules of differentiation. Differentiation of \(x\) gives \(1\) since the derivative of \(x\) concerning \(x\) is simply \(1\). Meanwhile, the derivative of \(2\sin x\) utilizes the differentiation of trigonometric functions, resulting in \(2\cos x\).
Thus, the full derivative is \(f'(x) = 1 + 2\cos x\). This equation helps us find where the function's graph has horizontal tangents by setting the derivative equal to zero and solving for \(x\).

Trigonometric functions, such as \(\sin x\) and \(\cos x\), are integral to solving various mathematical problems, particularly those involving periodic phenomena. These functions describe relationships in right-angled triangles and appear frequently in calculus when dealing with periodic functions.
In the given problem, the function \(f(x) = x + 2 \sin x\) includes a trigonometric component, \(2\sin x\). When differentiating this function, a trigonometric identity helps us establish the form of the derivative. The derivative of \(\sin x\) is \(\cos x\), which aids in finding \(f'(x) = 1 + 2\cos x\).
Solving for \(f'(x) = 0\) requires understanding the behavior and values of cosine, especially when \(\cos x\) gives specific solutions within trigonometric functions' known ranges in relation to the unit circle.

The unit circle is a valuable tool in trigonometry to visualize and understand the values of trigonometric functions at various angles. It helps determine the sine and cosine values for all angles in a proportional manner.
In the context of the exercise, the unit circle is essential in solving the equation \(\cos x = -\frac{1}{2}\). By referring to the coordinates on the unit circle, we find that \(\cos x = -\frac{1}{2}\) at angles \(x = \frac{2\pi}{3}\) and \(x = \frac{4\pi}{3}\).
This is because the x-coordinates on the unit circle are equal to the cosine values of angles. The problem further considers periodicity by adding \(2n\pi\), where \(n\) is an integer, to capture all possible angles that equate to the described cosine values.