Results 1 to 3 of 3

- April 30th 2007, 12:02 PM #1

- Joined
- Mar 2007
- Posts
- 32

## Just one More :)

Ok..maybe not one more but three

I'm supposed to find all solution of the equation in the inverval [0, 2pi).

1. sin2x = - sqrt(3)/2.

I tried solving this one. I have something like this: 2pi/3 + npi and 5pi/12 + npi.

2. tan3x = 1

3. Sec 4x = 2

Thanks for all the help!

Isabel

- April 30th 2007, 12:35 PM #2

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- April 30th 2007, 12:40 PM #3

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 12,028
- Thanks
- 846

Hello, Isabel!

Don't forget the interval . . .

Find all solutions in the inverval [0, 2π)

. . . . . . . . . . . . _

1) .sin(2x) .= .-√3/2.

We know that: .sin(π/3) = √3/2

. . and sine is negative in quadrants 3 and 4.

Hence: .2x .= .4π/3, 5π/3, 10π/3, 11π/3

Therefore: .x .= .2π/3, 5π/6, 5π/3, 11π/6

2) .tan(3x) .= .1

. . and tangent is positive in quadrants 1 and 3.

Hence: .3x .= .π/4, 5π/4, 9π/4, 13π/4, 17π/4, 21π/4

Therefore: .x .= .π/12, 5π/12, 3π/4, 13π/12, 17π/12, 7π/4

3) .sec(4x) .= .2

We know that: .sec(π/3) = 2

. . and secant is positive in quadrants 1 and 4.

Hence: .4x .= .π/3, 5π/3, 7π/3, 11π/3, 13π/3, 17π/3, 19π/3, 23π/3

Therefore: .x .= .π/12, 5π/12, 7π/12, 11π/12, 13π/12, 17π/12, 19π/12, 23π/12