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-(1-cos^2x)

2 cos^2 x+cosx=2cos^2x-1

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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