Trigonometric identities maths question please help
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Trigonometric identities maths question please help

[From: ] [author: ] [Date: 11-10-06] [Hit: ]
Let t = tan(X/2),when 1 - t² = -3hence t² = 4t = ±2X/2 = 63.435° or 116.henceX = 126.87° or 233.13°,......
The Question is: solve giving values of X from 0 to 360 degrees inclusive

2tan(X/2) + 3tan(X) = 0

Ive tried everything i can think of but nothing seems to work, please explain how you got the answer as well, thanks

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tan(X) = 2 tan(X/2) / (1 - tan²(X/2))

Let t = tan(X/2), and the equation becomes
2t + 3*2t/(1 - t²) = 0
2t[1 + 3/(1-t²)] = 0
therefore
t = 0 or 1 + 3/(1 - t²) = 0

t = 0 when X/2 = 0° or 180° hence X = 0° or 360°
3/(1 - t²) = -1
when 1 - t² = -3
hence t² = 4
t = ±2
X/2 = 63.435° or 116.565°
hence
X = 126.87° or 233.13°, together with the values 0° and 360° found above

After seeing brasshopper's answer, I too went to wolframalpha and typed in
2 tan(x/2) + 3 tan(x) zeros
and it came up with an exact general expression for all zeros. However it did also show the graph with three of the zeros, in radians (not degrees). One is zero. The only positive one shown is 2.2, which converts to about 126 degrees, so I'd say this confirms the answers I've found.

If still in any doubt, you can confirm my answers by substituting each of them in the original left-hand side of the equation. All of them will give 0, the right-hand side.

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I ran it through a numerical solver, and it came back with x=0. I then tried graphing the relations separately and they seemed to meet at zero and every 2pi. Finally I graphed them together and they graphed to a horizontal line.

Could be that they just don't have a solution.

Wolframalpha shows roots at zero and -2.21, see ref 1
1
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