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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? 

 
Julius  Dec 5, 2017

Best Answer 

 #2
avatar+79741 
+3

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

 

 

cool cool cool

 
CPhill  Dec 5, 2017
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6+0 Answers

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

 
Guest Dec 5, 2017
 #2
avatar+79741 
+3
Best Answer

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

 

 

cool cool cool

 
CPhill  Dec 5, 2017
 #3
avatar+357 
+1

Ahhh yes!! What was I thinking, that makes sense. 

 

Thanks 

 
Julius  Dec 5, 2017
 #4
avatar+357 
+1

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
 #5
avatar+79741 
+1

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

 

Thanks for noticing this  !!!

 

 

cool cool cool

 
CPhill  Dec 5, 2017
edited by CPhill  Dec 5, 2017
 #6
avatar+357 
+1

No problem :D

 
Julius  Dec 5, 2017

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