# Precalculus

posted by on .

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 - ,

Have you tried the double angle formulas?

• Precalculus - ,

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 - ,

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 - ,

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 - ,