Question
Find the general solution of : $\sin x \tan x=\tan x-\sin x+1$.
✓
Answer
sinx · tanx = tanx - sinx + 1
$\sin x \cdot\left(\frac{\sin x}{\cos x}\right)-1=\frac{\sin x}{\cos x}-\sin x$
$\frac{\sin ^2 x-\cos x}{\cos x}=\frac{\sin x-\sin x \cdot \cos x}{\cos x}$
$\sin ^2 x-\cos x=\sin x-\sin x \cdot \cos x$
now $x$ can't equal to $\frac{\pi}{2}$
$\sin ^2 x-\cos x-\sin x+\sin x \cos x=0$
$\sin ^2 x+\sin x \cos x-\cos x-\sin x=0$
$\sin x(\sin x+\cos x)-1(\sin x+\cos x)=0$
$(\sin x-1)(\sin x+\cos x)=0$
since $x$ is not $\frac{\pi}{2}$
$\sin x+\cos x=0$
$\sin x=-\cos x$
$x=$ multiple of $\frac{\pi}{4}$ in even quads.
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