Question

Find the domain of the function. $$ h(x)=\frac{1}{\sin x-\frac{1}{2}} $$

Short answer

The domain of the function \( h(x) = \frac{1}{\sin x-\frac{1}{2}} \) is all real numbers except \(x = \frac{\pi}{6} + 2\pi k\) and \(x = \frac{5\pi}{6} + 2\pi k\), where k is an integer.

Step-by-step solution

Identify the Critical Point

To solve this problem, first set the denominator of the function equal to zero and solve the equation for x. Thus, the equation will be \(\sin x-\frac{1}{2} = 0\)

Solve for x

Re-arrange and solve for x, the equation now becomes \(\sin x = \frac{1}{2}\). Solve this equation for all possible solutions. The solutions are \(x = \frac{\pi}{6} + 2\pi k\) and \(x = \frac{5\pi}{6} + 2\pi k\), where k is an integer.

Define the Function Domain

The solution in step 2 represents the values at which the denominator is zero, and thus, these values need to be excluded from the domain of the function \(h(x)\). Therefore, the function domain will be all real numbers except \(x = \frac{\pi}{6} + 2\pi k\) and \(x = \frac{5\pi}{6} + 2\pi k\), where k is an integer.