Question 925954: Find the x-intercepts algebraically
csc^2x-4cotx=-2
1/sin^2-4cosx/sinx=-2
1/sin^2x-4cosxsinx/sin^2x=-2
(1-4cosxsinx)/sin^2=-2
1-4cosxsinx=-2sin^2x
1-2sin2x=-2((1-cos2x)/2)
1-2sin2x=-(1-cos(2x))
This is what I've tried, but I can't figure out what comes next Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Find the x-intercepts algebraically
csc^2x-4cotx=-2
1/sin^2-4cosx/sinx=-2
1/sin^2x-4cosxsinx/sin^2x=-2
(1-4cosxsinx)/sin^2=-2
1-4cosxsinx=-2sin^2x
1-2sin2x=-2((1-cos2x)/2)
1-2sin2x=-(1-cos(2x))
-------
1-2sin(2x)= cos(2x) - 1
2 - 2sin(2x) = cos(2x) = sqrt(1 - sin^2(2x))
square both sides
4 - 8sin(2x) + 4sin^2(2x) = 1 - sin^2(2x)
Now it's a quadratic in 2x
----
(I took your word on the steps you did)
