Question

\\\frac{2\cos x + 2\cos^2 x}{\sin 2x} = \frac{\sin x}{\cos x - 1}\\

Explanation:

Identify the equation and domain restrictions

We are given the trigonometric equation:

$$ \frac{2\cos x + 2\cos^2 x}{\sin 2x} = \frac{\sin x}{\cos x - 1} $$

First, we must identify the values of \(x\) for which the expressions are defined. The denominators cannot be zero:

1. \(\sin 2x

eq 0 \implies 2x
eq k\pi \implies x
eq \frac{k\pi}{2}\) for any integer \(k\). This means \(\sin x
eq 0\) and \(\cos x
eq 0\).

1. \(\cos x - 1

eq 0 \implies \cos x
eq 1 \implies x
eq 2k\pi\).

Simplify the left-hand side

Using the double-angle identity \(\sin 2x = 2\sin x \cos x\), we rewrite the denominator of the left-hand side:

$$ \frac{2\cos x(1 + \cos x)}{2\sin x \cos x} $$

Since \(\cos x
eq 0\), we can cancel the common terms \(2\cos x\) from the numerator and the denominator:

$$ \frac{1 + \cos x}{\sin x} $$

Equate the simplified sides

Now, substitute the simplified left-hand side back into the original equation:

$$ \frac{1 + \cos x}{\sin x} = \frac{\sin x}{\cos x - 1} $$

Cross-multiply to solve for \(x\):

$$ (1 + \cos x)(\cos x - 1) = \sin^2 x $$

$$ \cos^2 x - 1 = \sin^2 x $$

Solve the simplified equation

Using the Pythagorean identity \(\sin^2 x = 1 - \cos^2 x\), we substitute this into the equation:

$$ \cos^2 x - 1 = 1 - \cos^2 x $$

Rearranging the terms gives:

$$ 2\cos^2 x = 2 $$

$$ \cos^2 x = 1 $$

This yields two possible sets of solutions:

$$ \cos x = 1 \quad \text{or} \quad \cos x = -1 $$

Check against domain restrictions

Let's analyze the solutions:

1. If \(\cos x = 1\), this violates the restriction \(\cos x - 1

eq 0\) (the denominator on the right-hand side would be zero). Thus, \(\cos x = 1\) is extraneous.

1. If \(\cos x = -1\), then \(\sin x = 0\). However, our domain restrictions require \(\sin 2x

eq 0\), which means \(\sin x
eq 0\). Thus, \(\cos x = -1\) is also extraneous because it makes the denominator \(\sin 2x = 2\sin x\cos x = 0\).

Since all potential solutions are extraneous, the equation has no solution.

Answer:

No solution