Question

2 sin x cos x = tan x​

Answer 1

If
`sin x = 0` we get
`2 (0) (\pm1) = 0` which is true. so we have a first set of solutions given by

`x= 0; x = \pi`;

Else, if
`sin x \\e 0` we can divide both sides by it and we get

`2 cos x = \frac 1 {cos x} \rightarrow 2cos^2x =1 \rightarrow cos^2x =\frac12\\cos x= \pm \frac{\sqrt2}2`

Which gives us a second set of solutions given by
`x= \frac{\pi}4;x= \frac{3\pi}4; x= \frac{5\pi}4;x= \frac{7\pi}4`;

We can group all solutions (doesn't matter, but it's more elegant!) by writing

`x= k\frac{\pi}4; k\in \{0,1,3,4,5,7\}`