Question

Precalculus @Hero

Answer 1

\[\sin x \tan x - \sin x = 0\]

Restriction: \[[-\frac{ \pi }{ 2 }, \frac{ \pi }{ 2 }]\]

\[\sin x \tan x = \sin x\]
\[\tan x = 1\]
\[x = \frac{ \pi }{ 4 }\]

This is what I get. But apparently it's also zero. What am I missing?

Answer 2

You factor out the sin(x) and use the zero product property to find two solutions.

Answer 3

oh

Answer 4

sin(x) = 0 and tan(x) - 1 = 0

Answer 5

\[\sin x ( \tan x -1) = 0\]
\[\sin x = 0\]
\[\tan x = 1\]

Answer 6

I see, thanks Hero

Answer 7

Exactly