anonymous
  • anonymous
Solve Trig Equation \( tan \theta = \frac{2\sqrt{3}}{3} sin \theta \)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I need to get the solution sets
imqwerty
  • imqwerty
hint u can write tan(theta) as sin(theta)/cos(theta)
anonymous
  • anonymous
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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anonymous
  • anonymous
|dw:1433617958179:dw|
imqwerty
  • imqwerty
nw cancel sin(theta) from both sides of the equation
imqwerty
  • imqwerty
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
anonymous
  • anonymous
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.
imqwerty
  • imqwerty
cos(theta) = cos(30) = cos(330)
anonymous
  • anonymous
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.
anonymous
  • anonymous
I am stuck on factoring the last part
anonymous
  • anonymous
\[\tan(0) = \frac{ 2\sqrt{3} }{ 3 }\sin(0)\] YOu get 0 = 0. You lose this solution if you cancel out sin(theta) from both sides.
imqwerty
  • imqwerty
ok m posting a full solution
anonymous
  • anonymous
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 \[\frac{ 2\sqrt{3} }{ 3 }\cos\theta -1 = 0\]
anonymous
  • anonymous
|dw:1433618602641:dw|
anonymous
  • anonymous
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
anonymous
  • anonymous
Where does the -1 come from? Is it because cos -sin = cos - 1 sin ???
anonymous
  • anonymous
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.
anonymous
  • anonymous
Ok I see thanks
anonymous
  • anonymous
But from above, I have solutions that will come from \[\sin\theta = 0\] and \[\frac{ 2\sqrt{3} }{ 3 }\cos\theta - 1 = 0\]
imqwerty
  • imqwerty
√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
anonymous
  • anonymous
I think I got it. Thanks to you all
anonymous
  • anonymous
You're welcome :)
anonymous
  • anonymous
Yep I got it. My solutions were 0, pi, 5pi/6, 7pi/6 Thanks once again to you all.
imqwerty
  • imqwerty
:)

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