Question 1155366
The first thing is to look for zeros:
tan(x)=0 at  0, 180, and 360 degrees, and these are all solutions.
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For non-zero solutions:

Simplify  {{{2sin(x)tan(x) = -tan(x) }}} to {{{ sin(x) = -1/2 }}}, {{{tan(x)<>0}}}
and recall:
{{{sin(30^o) = 1/2 }}} and
in Q1,Q2,Q3,Q4, sin(x) is +,+,-,-, respectively.<br>

Working from the x-axis, push the angle into Q3 by adding 180+30 = 210 and into Q4 by subtracting 30 from 360, giving {{{cross(310)}}}330  (edited typo here).<br>

Solutions are {{{ highlight(matrix(1,9,0^o,",",180^o,",",210^o,",",330^o,",and" ,360^o)) }}}

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A picture is worth a thousand words:
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*[illustration sin_tan]

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And note if you rearrange to 2sin(x)tan(x)+tan(x) = 0  and graph the LHS, you can simply look for where this rearranged function crosses the x-axis:

*[illustration sin_tan_p_tan_eq0]