A student attempts to solve the equation

in the range $0\le x\le 2\pi$.

The student's attempt is as follows:

$\cos x+\sin x\tan x=2\sin x-1$

So $\cos x-\sin x+\sin x\tan x-\sin x=-1$ (I)

So $(\sin x-\cos x)(\tan x-1)=-1$ (II)

So $\sin x-\cos x=-1\text{ or }\tan x-1=-1$ (III)

So $(\sin x-\cos x{)}^2=1\text{ or }\tan x=0$ (IV)

So $2\sin x\cos x=0\text{ or }\tan x=0$ (V)

So $x=0,\frac{\pi }{2},\pi ,\frac{3\pi }{2},2\pi$ (VI)

Which of the following best describes this attempt?