 # Need help with trigonometry and intervals !! Anonymous
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### 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   = ?

nicolasv09
School: UCLA   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

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.

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

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)

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Thanks, good work Brown University

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