Lizbeth,
Starting with 4 sin(4x)= -8 sin(2x), let's let u=2x. Then we have
4 sin(2u)= -8 sin(u). Add -8sin(u) to both sides to get: 4 sin(2u)+8 sin(u)=0.
Our trig identities tell us that sin 2θ = 2sinθcosθ. Substituting that in our equation, we have
4• 2sin(u) cos(u)+8 sin(u) = 0, or
8 sin(u) cos(u)+ 8 sin(u)=0.
Factor the expression next:
8 sin(u) (cos(u) +1)=0.
Set each factor = 0 to solve for u:
8 sin(u)=0
sin (u)=0.
Between 0 and 2π, the only places where sin(u)=0 are at u=0 and u=π, since those are both on the x-axis, where y=0.
The next factor is (cos(u) + 1):
cos(u) +1 = 0.
Subtract 1 from both sides to get
cos(u) = -1.
On the unit circle, x= -1 at u=π.
Now, that's all the values for u. u=0,π.
Woot!
...but we're looking for x, not u. Recall that u=2x. Plug the values of u in that equation to get x:
0=2x gives us x=0.
π=2x gives us x=π/2.
Since trig functions measure things that *cycle*, we can find more values of x from the second answer.
For π=2x, going around the unit circle n times gives us an infinite number of answers by adding (2π•n)/2 to the answer. Within the domain of [0,2π), only a few fit.
Given n=1: x=π/2 + 2π/2 = 3π/2.
Given n=2: x=π/2 + (2π•2)/2 = π/2 + 4π/2 = 5π/2, which is greater than 2π and outside of our domain. This means we've got all our answers.
x= 0, π/2, 3π/2.
I hope this isn't too confusing. It's a long process, but it's all stuff you know how to do! Factoring, reading your unit circle, and solving equations.
Happy solving!
(and let me know if you have any questions)
Liz Z.
