Need help with trigonometry and intervals !!

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1)    solve the equation on the interval, [0,2pi) .

6 tan^2 x -17 tanx +7 =0


2)  solve the equation on the interval, [0,2pi)

sinx = -2/3

3) 2 tan^2x -2 =0

4) olve the equation on the interval, [0,2pi)

   (cot u +1)(csc u -1) = 0

   u   = ?

Nov 24th, 2015

Thank you for the opportunity to help you with your question!

1. Think of the equation as a quadratic, substitute x instead of tan(x).

6 x^2 - 17 x + 7 = 0. (given it is a quadratic we can factor it)

(3 x  - 7) (2x -1) = 0 (but since the original uses tan instead of x, we substitute tan(X) back in)

(3 tan (x) - 7) ( 2 tan(x) -1) = 0 . (now we solve for x for each equation)

3 tan (x) = 7 which is tan(x) = 7/3    and 2 tan(x) = 1   which is tan(x) = 1/2

Using a calculator we take the arc tangent of the values (in radian mode) and our answer is 

1.166 radians and 0.46364 radians  .

BUT! Since tangent is positive in both 1st and 3rd quadrant, there are 2 answers for each. To find the answer, we have to add Pi radians to both answers, or 3.1415.

So the 4 answers are: 1.166 radians, 4.3075 radians, 0.4364 radians, and 3.578 radians


2. sin (x) = -2/3

using a graphical approach, we know there are 2 answer, one in the 3rd and one in the 4th quadrant

using arc sin(X), we get the value to be -0.7297 radians. Since its from 0 to 2pi, we add the value (since it is negative) to 2 pi.

6.2832 radians + -0.7297 radians, gives us: 5.553 radians

Since the other answer is in the 3rd quadrant, what we do is add the 0.7297 value to 1pi to get the 3rd quadrant answer. So:

3.1415 radians + 0.7297 radians gives us: 3.8713 radians


3. solve so tan(2x) is by itself. 

tan 2x = 1

tan (x) = 1 at pi/4 and 5pi/4. so then we have to find half of those respective values

meaning your answers are pi/8 and 5pi/8


4. 2 answers

cot (u) = -1

cot(u) = -1 when sin and cos are the same, except one is positive and the other negative, meaning 2nd and 4th quadrant. That is true at 3pi/4 and 7pi/4.


csc u = 1. csc is the same thing as 1/sin. So for that to be true, sine must also be 1, because 1/1 = 1.

That is only true at 1 point, (pi/2)





Please let me know if you need any clarification. I'm always happy to answer your questions.
Nov 24th, 2015

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