Find all solutions for 2 sin^2(u) + sin(u) = 1, etc.

[dewykitten13]

I have three problems for find all solutions in the interval [0, 2pie).
1) 2sin[sup:3odc9kmo]2[/sup:3odc9kmo]u + sin u =1
I moved the one over and factored...and set the two part equal to zero seperately. I got sin u=-1 and sin u= 1/2.
I got 5 solutions...pie/2, pie/6, 2pie/3, 7pie/6, and 5pie/3. Are these correct?
2) sin3xcosx + cos3xsinx = -1/2
I factored out a sinx and cos x and got sinxcosx(sin2x+cos2x)=-1/2
I set each one equal to -1/2 seperately and got sinx=-1/2 cosx=-1/2 and sin2x+cos2x=-1/2
This is where I got stuck
3) 2- cos[sup:3odc9kmo]2[/sup:3odc9kmo]x=4sin[sup:3odc9kmo]2[/sup:3odc9kmo](x/2)
I am completely lost here

 

[skeeter]

Re: Find all solutions

I have three problems for find all solutions in the interval [0, 2pie).

btw ... it's spelled pi, not "pie". pie is something you eat.

1) 2sin[sup:1sf8416g]2[/sup:1sf8416g]u + sin u =1
I moved the one over and factored...and set the two part equal to zero seperately. I got sin u=-1 and sin u= 1/2.
I got 5 solutions...pie/2, pie/6, 2pie/3, 7pie/6, and 5pie/3. Are these correct?

no, you have some invalid solutions. sin(u) = -1 at u = 3pi/2 ... sin(u) = 1/2 at u = pi/6 and u = 5pi/6

2) sin3xcosx + cos3xsinx = -1/2
I factored out a sinx and cos x and got sinxcosx(sin2x+cos2x)=-1/2
I set each one equal to -1/2 seperately and got sinx=-1/2 cosx=-1/2 and sin2x+cos2x=-1/2
This is where I got stuck

hint ... sin(a+b) = sin(a)cos(b) + cos(a)sin(b)

3) 2- cos[sup:1sf8416g]2[/sup:1sf8416g]x=4sin[sup:1sf8416g]2[/sup:1sf8416g](x/2)
I am completely lost here

another hint ... sin[sup:1sf8416g]2[/sup:1sf8416g](x/2) = [1 - cos(x)]/2