kinuel8091
New member
- Joined
- Jul 27, 2013
- Messages
- 24
Solve the equation for solutions in the interval 0 ≤ x < 2π. Round approximate solutions to the nearest ten-thousandth.
sec2(x)+√3 sec x-√2 sec x -√6=0
(sec (x)+√3)(sec x -√2)=0
sec(x)-√2=0
sec(x)=√2
x=arcsec √2
x=π/4 , 7π/4
sec (x)+√3=0
sec(x)=-√3
x=arcsec (-√3)
x=????
arcsec (-√3)=arccos (-1/√3)
right?
So I typed in my calculator which is in "DEG".
arccos (-1/√3)
and the result was =125.2643897
For this one, i have to consider the 2nd quadrant angle ang 3rd quadrant angle right?
x=125.2643897
ref angle = 180-125.3 =54.7
3rd quadrant angle=180+54.7=234.7
or should it be in "RAD"
which is equal to=2.186276035
same as throug with this one
pi-2.2=0.9
pi+0.9=4.0
which one should i input here if the range is in radians?
sec2(x)+√3 sec x-√2 sec x -√6=0
(sec (x)+√3)(sec x -√2)=0
sec(x)-√2=0
sec(x)=√2
x=arcsec √2
x=π/4 , 7π/4
sec (x)+√3=0
sec(x)=-√3
x=arcsec (-√3)
x=????
arcsec (-√3)=arccos (-1/√3)
right?
So I typed in my calculator which is in "DEG".
arccos (-1/√3)
and the result was =125.2643897
For this one, i have to consider the 2nd quadrant angle ang 3rd quadrant angle right?
x=125.2643897
ref angle = 180-125.3 =54.7
3rd quadrant angle=180+54.7=234.7
or should it be in "RAD"
which is equal to=2.186276035
same as throug with this one
pi-2.2=0.9
pi+0.9=4.0
which one should i input here if the range is in radians?