Answer
$0$, $\frac{\pi}{4}$, $\frac{5\pi}{4}$
Work Step by Step
$\sin x-\cos x-\tan x=-1$
$\sin x-\cos x-\frac{\sin x}{\cos x}=-1$
Multiply both sides by $\cos x$:
$\sin x\cos x-\cos^2 x-\sin x=-\cos x$
$\cos x-\cos^2 x-\sin x+\sin x\cos x=0$
$\cos x(1-\cos x)-\sin x(1-\cos x)=0$
$(\cos x-\sin x)(1-\cos x)=0$
If $\cos x-\sin x=0$, then $\cos x=\sin x$. Dividing both sides by $\cos x$ gives us $\tan x=1$. The only solutions in $[0, 2\pi)$ are $\frac{\pi}{4}$ and $\frac{5\pi}{4}$.
If $1-\cos x=0$, then $1=\cos x$. The only solution in $[0, 2\pi)$ is $0$.
