# Need help with trigonometry and intervals !!

*label*Mathematics

*timer*Asked: Nov 24th, 2015

*account_balance_wallet*$5

### Question Description

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 = ?

## Tutor Answer

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.

*flag*Report DMCA

Brown University

1271 Tutors

California Institute of Technology

2131 Tutors

Carnegie Mellon University

982 Tutors

Columbia University

1256 Tutors

Dartmouth University

2113 Tutors

Emory University

2279 Tutors

Harvard University

599 Tutors

Massachusetts Institute of Technology

2319 Tutors

New York University

1645 Tutors

Notre Dam University

1911 Tutors

Oklahoma University

2122 Tutors

Pennsylvania State University

932 Tutors

Princeton University

1211 Tutors

Stanford University

983 Tutors

University of California

1282 Tutors

Oxford University

123 Tutors

Yale University

2325 Tutors