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!
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
We know that: .tan(π/4) = 12) .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