Posted by Anonymous on Thursday, January 3, 2013 at 5:14pm.
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

Precalculus  bobpursley, Thursday, January 3, 2013 at 5:40pm
Have you tried the double angle formulas?

Precalculus  Anonymous, Thursday, January 3, 2013 at 5:58pm
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.

Precalculus  bobpursley, Thursday, January 3, 2013 at 6:07pm
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) = cosĀ²(x)  sinĀ²(x)
2 cos^2 x+cosx=cos^2x(1cos^2x)
2 cos^2 x+cosx=2cos^2x1
cosx=1
x=PI

Precalculus  Steve, Thursday, January 3, 2013 at 6:44pm
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

Precalculus  Anonymous, Thursday, January 3, 2013 at 9:33pm
Thanks for your help you guys.
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