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# Solving trigonomerty equation

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Solve : tg(x)>=sin(x) on  interval <0,2pi>

Anybody who know solve this?  becuase im lost

blaster0  May 18, 2014

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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.

Alan  May 18, 2014
#1
+27229
+10

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.

Alan  May 18, 2014
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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

blaster0  May 18, 2014
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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

Alan  May 18, 2014
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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.

Melody  May 19, 2014
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You are right Melody.  I really must stop doing stuff late at night!

Alan  May 19, 2014
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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.

CPhill  May 19, 2014