A student attempts to solve the equation $$\cos x + \sin x \tan x = 2 \sin x - 1$$ in the range $0 \leq x \leq 2 \pi$. The student's attempt is as follows: $\cos x + \sin x \tan x = 2 \sin x - 1$ So $\quad \cos x - \sin x + \sin x \tan x - \sin x = - 1$ So $\quad ( \sin x - \cos x ) ( \tan x - 1 ) = - 1$ So $\quad \sin x - \cos x = - 1$ or $\tan x - 1 = - 1$ So $\quad ( \sin x - \cos x ) ^ { 2 } = 1$ or $\tan x = 0$ So $\quad 2 \sin x \cos x = 0$ or tan $x = 0$ So $x = 0 , \frac { \pi } { 2 } , \pi , \frac { 3 \pi } { 2 } , 2 \pi$ Which of the following best describes this attempt?
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A student attempts to solve the equation
$$\cos x + \sin x \tan x = 2 \sin x - 1$$
in the range $0 \leq x \leq 2 \pi$.
The student's attempt is as follows:
$\cos x + \sin x \tan x = 2 \sin x - 1$
So $\quad \cos x - \sin x + \sin x \tan x - \sin x = - 1$
So $\quad ( \sin x - \cos x ) ( \tan x - 1 ) = - 1$
So $\quad \sin x - \cos x = - 1$ or $\tan x - 1 = - 1$
So $\quad ( \sin x - \cos x ) ^ { 2 } = 1$ or $\tan x = 0$
So $\quad 2 \sin x \cos x = 0$ or tan $x = 0$
So $x = 0 , \frac { \pi } { 2 } , \pi , \frac { 3 \pi } { 2 } , 2 \pi$
Which of the following best describes this attempt?