Solve for \( x \):
[tex]\[ 3x = 6x - 2 \][/tex]



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[tex]\[ 1-\frac{\sin x}{\tan x}=\cos x \][/tex]
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Response:
[tex]\[ 1 - \frac{\sin x}{\tan x} = \cos x \][/tex]



Answer :

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.

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