Let's solve the equation:
[tex]\[ 1 - \frac{\sin x}{\tan x} = \cos x. \][/tex]
First, recall the trigonometric identity for the tangent function:
[tex]\[ \tan x = \frac{\sin x}{\cos x}. \][/tex]
Substitute this identity into the equation:
[tex]\[ 1 - \frac{\sin x}{\frac{\sin x}{\cos x}} = \cos x. \][/tex]
Simplify the fraction:
[tex]\[ 1 - \frac{\sin x \cdot \cos x}{\sin x} = \cos x. \][/tex]
Notice that \(\sin x\) in the numerator and denominator cancels out:
[tex]\[ 1 - \cos x = \cos x. \][/tex]
Now we have a simpler equation:
[tex]\[ 1 = 2\cos x. \][/tex]
Solve for \(\cos x\) by isolating it on one side:
[tex]\[ \cos x = \frac{1}{2}. \][/tex]
Next, determine the values of \(x\) that satisfy \(\cos x = \frac{1}{2}\). The cosine function equals \(\frac{1}{2}\) at specific angles within one full revolution (from 0 to \(2\pi\)). These angles are:
[tex]\[ x = \pm \frac{\pi}{3} + 2k\pi, \][/tex]
where \(k\) is any integer.
These angles can be specified more explicitly within the range \([-\pi, \pi]\) as:
[tex]\[ x = \frac{\pi}{3}, \; x = -\frac{\pi}{3}, \; x = \frac{5\pi}{3}, \; x = -\frac{5\pi}{3}. \][/tex]
Thus, the solution to the given equation \( 1 - \frac{\sin x}{\tan x} = \cos x \) is:
[tex]\[ x = -\frac{5\pi}{3}, -\frac{\pi}{3}, \frac{\pi}{3}, \frac{5\pi}{3}. \][/tex]
These are the values of [tex]\(x\)[/tex] that satisfy the given equation.
