Trig Functions — Question 7

Question 7 - Solution

Simplify. Where the original function is defined,

$$f(x)=\frac{2\cos^2x}{2\sin x\cos x}=\cot x.$$

The cancellation does not restore points where $\sin(2x)=0$.

Solve. On $(0,\pi)$, $\cot x=2$ gives

$$\boxed{x=\arctan(1/2)\approx0.4636}.$$

This point is in the original domain.

Domain. All multiples of $\pi/2$ must be excluded. On $[0,2\pi]$ the domain is

$$\boxed{(0,\pi/2)\cup(\pi/2,\pi)\cup(\pi,3\pi/2)\cup(3\pi/2,2\pi)}.$$