Solve Trig Equation \( tan \theta = \frac{2\sqrt{3}}{3} sin \theta \)

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Solve Trig Equation \( tan \theta = \frac{2\sqrt{3}}{3} sin \theta \)

Mathematics
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I need to get the solution sets
hint u can write tan(theta) as sin(theta)/cos(theta)
Well, first off we should note that we cannot have solutions that would make cosine equal to 0, as then we would have something undefined. So let me change tangent into sin/cos and do this: \[\frac{ \sin\theta }{ \cos\theta } = \frac{ 2\sqrt{3} }{ 3 }\sin\theta\] \[\sin\theta = \frac{ 2\sqrt{3} }{ 3 }\sin\theta\cos\theta\] \[\frac{ 2\sqrt{3} }{ 3 }\sin\theta\cos\theta - \sin\theta = 0\] Now factor and see if you solve from there

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|dw:1433617958179:dw|
nw cancel sin(theta) from both sides of the equation
then u'll b left with 1/cos(theta) = 2/root(3) cross multiplying we get cos(theta) = root(3)/2 which is equal to cos(30) therefore theta = 30degrees
In general, you do not want to cancel sin(theta) from both sides of the equation, or divide by any variable expression for that manner. If you do this, you could potentially lose solutions.
cos(theta) = cos(30) = cos(330)
To show this, by inspection you can see that 0 would be a solution to this equation. Yet if you divide out sin(theta), you completely lose 0 as a solution in what would be remaining.
I am stuck on factoring the last part
\[\tan(0) = \frac{ 2\sqrt{3} }{ 3 }\sin(0)\] YOu get 0 = 0. You lose this solution if you cancel out sin(theta) from both sides.
ok m posting a full solution
As for the factoring, where I have \[\frac{ 2\sqrt{3} }{ 3 }\sin\theta\cos\theta - \sin\theta = 0\] you can factor out sin(theta) \[\sin \theta(\frac{ 2\sqrt{3} }{ 3 }\cos\theta -1) = 0\] This gives two possibilities. \[\sin\theta = 0\] or \[\frac{ 2\sqrt{3} }{ 3 }\cos\theta -1 = 0\]
|dw:1433618602641:dw|
You cannot divide by sin(theta). In general, you cannot divide by variable expression. Notice that I had one of my potential results when I factored as sin(theta) = 0. You completely lose this solution if you divide out sin(theta), you are not allowed to do this @Nixy
Where does the -1 come from? Is it because cos -sin = cos - 1 sin ???
As a small example of why the -1, consider 2x-2. I can factor a 2 out of this expression and get 2(x-1). When you factor a number out of itself, it leaves behind a 1. You can also distribute to check this result if you wish.
Ok I see thanks
But from above, I have solutions that will come from \[\sin\theta = 0\] and \[\frac{ 2\sqrt{3} }{ 3 }\cos\theta - 1 = 0\]
√3 tan x = 2 sin x √3 tan x - 2 sin x = 0 sin x ( √3/cos x - 2 ) = 0 => sin x = 0 or √3 / cos x - 2 = 0 when sin x = 0 x = 2kpi where k= 0,1,2,3......... when √3 / cos x - 2 = 0 => cos x = √3/2
I think I got it. Thanks to you all
You're welcome :)
Yep I got it. My solutions were 0, pi, 5pi/6, 7pi/6 Thanks once again to you all.
:)

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