Trig Functions — Question 8

Question 8 - Solution

Simplification and domain. The original function is defined for $x\ne\pi/2+2k\pi$. Where $\cos x\ne0$, rationalization gives

$$f(x)=\frac{1+\sin x}{\cos x}=\sec x+\tan x.$$

At $x=3\pi/2+2k\pi$, the original function has value $0$, although this alternative expression is undefined.

Solve on the requested interval. If $f(x)=1$, then $\cos x+\sin x=1$, with $\sin x\ne1$.

Using $\cos x=1-\sin x$ in $\cos^2x+\sin^2x=1$ gives

$$2\sin x(\sin x-1)=0.$$

The case $\sin x=1$ is excluded. The other case requires $\sin x=0$ and $\cos x=1$, hence $x=2k\pi$.

Neither endpoint belongs to $(0,2\pi)$, so

$$\boxed{\text{There are no solutions in }(0,2\pi).}$$

Parity. The domain is not symmetric about zero: $-\pi/2$ is allowed but $\pi/2$ is not.

$$\boxed{f\text{ is neither even nor odd}.}$$