Solve for x in the following equation.

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

This problem has an endless number of possible solutions. Real numbers that cause the denominators to equal zero must be excluded from the set of possible solutions since denominators of fractions cannot equal zero.

sin(x) ≠ 0 -> x ≠ 0 ± nπ

cos(x) ≠ -1 -> x ≠ π/2 ± nπ

The set of real numbers in the set {± nπ} must be excluded from the possible set of solutions.

There are three trigonometric terms in the equation.

Step 1:Multiple the second equation by 1 in the form $\frac{1 - cos(x)}{1 - cos(x)}$, simplify, and solve it.

Step 2:Check your answers with the original equation:

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

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

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

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

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

$\frac{2}{sin(x)} = \frac{2}{sin(x)}$

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