Solving trigonomerty equation

Are you sure the question is complete?  The graph below shows that all values in the region shaded in red (which extends up to infinity!) satisfies the condition ≥sin(x).

If tg(x) is restricted to the range 0 to 2pi as well, then it's all the red values that are shown above zero, stretching up to 2pi.

it so assigned:

Solve on interval <0;2*pi> non-equation: tgx>=sinx 

Here is result

http://www.wolframalpha.com/input/?i=tgx%3E%3Dsinx+%26%26+x%3E0%3C2pi

but i dont know how to solve it 

Ah! tan(x)≥sin(x)

Edit.

Melody is correct below!  The correct result is given when tan(x)-sin(x)>=0. i.e. when 0≤x<pi/2 and pi≤x<3pi/2 

y=tanx is in red,  y=sinx is in blue

$$tan(x)\ge sin(x)$$     when the red is equal or above the blue

solution:  

  $$\left[0\le x< \frac{\pi}{2}\right)\quad \bigcup \quad \left[\pi\le x< \frac{3\pi}{2}\right)$$

That is what I think anyway.

You are right Melody.  I really must stop doing stuff late at night! 

I believe that Melody is correct and we don't even need a graph to prove it.

Note that, in the first quadrant, the tangent is equal to the sine at 0, but at each angle until pi/2, the tangent is greater beause the sine function is being "divided" by 1 whlle the sine function in the tangent is being divided by a cosine function which is less than 1 (but greater than 0).

In quadrant two, the sine is positive and the tangent is negative.

This situation in quadrant 2 is reversed in quadrant 3. The tangent is positive and the sine is negative.  (They are only = at pi.)

In quadrant 4, the situation is reversed from the 1st quadrant...i.e...the sine is a smaller negative than the tangent.

Thus Melody's answer of [0, pi/2) U [pi, 3pi/2) seems good.