Please help!!!!!!!!!!! Find all solutions to the equation in the interval [0,2π).
8. cos2x=cosx
10. 2cos^2x+cosx=cos2x
Solve algebraically for exact solutions in the interval [0,2π). Use your grapher only to support your algebraic work.
24. cos2x+sinx=0
5 answers
Have you tried the double angle formulas?
Yes. It is just too hard and it would be nice if you would just explain it to me or show me an example. Please.
Hard? Why would anyone want to do something easy?
10.2cos^2x+cosx=cos2x
2cos^2x+cosx=cosx= cos(x)cos(x) - sin(x)sin(x) = cos²(x) - sin²(x)
2 cos^2 x+cosx=cos^2x-(1-cos^2x)
2 cos^2 x+cosx=2cos^2x-1
cosx=-1
x=PI
10.2cos^2x+cosx=cos2x
2cos^2x+cosx=cosx= cos(x)cos(x) - sin(x)sin(x) = cos²(x) - sin²(x)
2 cos^2 x+cosx=cos^2x-(1-cos^2x)
2 cos^2 x+cosx=2cos^2x-1
cosx=-1
x=PI
the only way to make these easy is to do dozens of them, not just the assignments. The double angle formulas are critical to know.
24.
cos 2x + sin x=0
1 - 2sin^2 x + sin x = 0
rearrange things a bit to the usual order:
2sin^2 x - sin x - 1 = 0
(2sin x + 1)(sin x - 1) = 0
so, either sin x = -1/2 or sin x = 1
sin x = -1/2: x = 7pi/6 or 11pi/6
sin x = 1: x = pi/2
24.
cos 2x + sin x=0
1 - 2sin^2 x + sin x = 0
rearrange things a bit to the usual order:
2sin^2 x - sin x - 1 = 0
(2sin x + 1)(sin x - 1) = 0
so, either sin x = -1/2 or sin x = 1
sin x = -1/2: x = 7pi/6 or 11pi/6
sin x = 1: x = pi/2
Thanks for your help you guys.
4 answers