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# Trigonometry?

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Hello,

I've been struggling a bit with where to start off from here. If someone could guide me that would be great!

Solve \(tan^2sinx -sinx/3\) = 0 on the interval 0 ≤ x ≤ 2π

The way I approached this was try to get rid of the fractions so I multiplied each side by 3. I then found my GCF of sinx and factored it through. It looked like this: sinx(3tan^2x-1)

Am I correct? Any ideas thoughts?

Dec 5, 2017

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tan^2 (x) * sin (x) - sin (x) / 3  = 0     get a common denominator

[ 3tan^2 (x) * sin (x) - sinx] / 3  = 0     mutiply both sides by 3

3tan^2 (x) * sin (x) - sin (x)  = 0        factor out sin(x)

sin (x)  [  3tan^2 ( x)  - 1]   =  0

We have two equations here.....either.......

sin (x)  = 0       and this happens at   0  and   pi

Or

3tan^2 (x)  -  1   =  0       add 1 to both sides

3tan^2 (x)   =   1           divide both sides by 3

tan^(2) x  =  1/3            take both roots

tan (x)  =  1/√3      and this happens at  pi/6   and 7pi/6

And

tan (x)  =  -  1/√3   and this happens at  5pi/6  and 11pi/6

So.....the solutions are    0, pi/6, 5pi/6, pi, 7pi/6 and 11pi/6   Dec 5, 2017

#1
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I think I woud divide through by sin x  to get started.....

Dec 5, 2017
#2
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tan^2 (x) * sin (x) - sin (x) / 3  = 0     get a common denominator

[ 3tan^2 (x) * sin (x) - sinx] / 3  = 0     mutiply both sides by 3

3tan^2 (x) * sin (x) - sin (x)  = 0        factor out sin(x)

sin (x)  [  3tan^2 ( x)  - 1]   =  0

We have two equations here.....either.......

sin (x)  = 0       and this happens at   0  and   pi

Or

3tan^2 (x)  -  1   =  0       add 1 to both sides

3tan^2 (x)   =   1           divide both sides by 3

tan^(2) x  =  1/3            take both roots

tan (x)  =  1/√3      and this happens at  pi/6   and 7pi/6

And

tan (x)  =  -  1/√3   and this happens at  5pi/6  and 11pi/6

So.....the solutions are    0, pi/6, 5pi/6, pi, 7pi/6 and 11pi/6   CPhill Dec 5, 2017
#3
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Ahhh yes!! What was I thinking, that makes sense.

Thanks

Julius  Dec 5, 2017
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Oh also you said sin x =0 at 0 and π, but sin x hits zero and 0, π and 2π. Correct?

Julius  Dec 5, 2017
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Yeah, Julius....I forgot about  the second  "≤" ....2pi is also a solution

Thanks for noticing this  !!!   Dec 5, 2017
edited by CPhill  Dec 5, 2017
#6
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No problem :D

Julius  Dec 5, 2017