I have three different equations that I have to find all the solutions to within the interval of [0, 2pi)
1. 3cos^2(x) - 5cos(x) = 4
I decided to move the four over and use the quadratic equation. so without the cosines, it looks like this:
x = ( 5+/- root(-5-4(3)(-4)))
______________________
2(3)
x = 5+root(43) or x= 5-root(43)
________ _________
6 6
the first one I got a domain error, but the second one i got that cosX=105.5. Using this, another answer could be 254.96. Is this correct?
2. 2(sin(x) + csc(x)) = 5
I can distribute the 2 so: 2sin(x) + 2csc(x) - 5 = 0
then change csc so:
2sin(x) + 1 -5 = 0
____
2sin(x)
but then I'm not sure how to solve from here.
3. 3cos2(x) + 8sin(x) + 5 = 0
I changed the cos2 into a double angle formula and simplified from there:
3(1-2sin^2(x)) + 8sinX + 5 = 0
3 - 6sin^2(x) + 8sin(x) + 5 = 0
-6sin^2(x) + 8sin(x) + 8 = 0
then I plugged it into the quadratic formula without the sines:
x= -8 +/- root(8^2 - 4(-6)(8))
______________________
2(-6)
x= -8 +/- root(256)
_____________
-12
x= -8 +/- 16
_______
-12
So, x could either be (-2/3) or (2).
using this, I found that sin(x) = -(2/3) was -41.8103, but that sin(x) = 2 was a domain error.
If I did the first one (-2/3) correctly, then could a possible answer between [0.2pi) be 318.19 (if you subtract 41.81 from 360) and also 221.81 (if you add 41.81 to 180) ?
thanks so much!
1. 3cos^2(x) - 5cos(x) = 4
I decided to move the four over and use the quadratic equation. so without the cosines, it looks like this:
x = ( 5+/- root(-5-4(3)(-4)))
______________________
2(3)
x = 5+root(43) or x= 5-root(43)
________ _________
6 6
the first one I got a domain error, but the second one i got that cosX=105.5. Using this, another answer could be 254.96. Is this correct?
2. 2(sin(x) + csc(x)) = 5
I can distribute the 2 so: 2sin(x) + 2csc(x) - 5 = 0
then change csc so:
2sin(x) + 1 -5 = 0
____
2sin(x)
but then I'm not sure how to solve from here.
3. 3cos2(x) + 8sin(x) + 5 = 0
I changed the cos2 into a double angle formula and simplified from there:
3(1-2sin^2(x)) + 8sinX + 5 = 0
3 - 6sin^2(x) + 8sin(x) + 5 = 0
-6sin^2(x) + 8sin(x) + 8 = 0
then I plugged it into the quadratic formula without the sines:
x= -8 +/- root(8^2 - 4(-6)(8))
______________________
2(-6)
x= -8 +/- root(256)
_____________
-12
x= -8 +/- 16
_______
-12
So, x could either be (-2/3) or (2).
using this, I found that sin(x) = -(2/3) was -41.8103, but that sin(x) = 2 was a domain error.
If I did the first one (-2/3) correctly, then could a possible answer between [0.2pi) be 318.19 (if you subtract 41.81 from 360) and also 221.81 (if you add 41.81 to 180) ?
thanks so much!