Solving 4sin3(x)=5sin(x) in [0,2╥)

[MacLaddy]

Homework Statement

4sin³(x)=5sin(x)

In the interval [0,2╥)

Homework Equations

Lot's and lot's of trig identities.

The Attempt at a Solution

4sin³(x)=5sin(x)
4sin³(x)-5sin(x)=0
sinx(4sin²(x)-5)=0

Setting sin(x)=0 gives me the solutions of 0, and ╥, but trying the other part
4sin²(x)-5=0
sin²(x)=⁵/₄
sin(x)=^√5/₂

This answer does not work at all for this solution, as my calculator just gives me constant errors when trying to find the inverse. I'm sure I'm going wrong somewhere on this.

Any help is appreciated.

 

[rock.freak667]

The Attempt at a Solution

4sin³(x)=5sin(x)
4sin³(x)-5sin(x)=0
sinx(4sin²(x)-5)=0

Setting sin(x)=0 gives me the solutions of 0, and ╥, but trying the other part

This is correct.

4sin²(x)-5=0
sin²(x)=⁵/₄
sin(x)=^√5/₂

This answer does not work at all for this solution, as my calculator just gives me constant errors when trying to find the inverse. I'm sure I'm going wrong somewhere on this.

Any help is appreciated.

Right so you have something like

sin(x)= 1.(something)

If you look at the graph of y=sin(x), you will see that -1≤sin(x)≤1

so your equation of sin(x) = √5/2 will lead to no real solutions.

 

[MacLaddy]

So if I understand you correctly, the other answers are extraneous, and only 0 and ╥ satisfy the equation?

Thank you for your help, I've been puzzling over this one for a couple of hours trying to figure that out.

 

[HallsofIvy]

I wouldn't use the term "extraneous" since that refers to roots of an equation produced while trying to solve an original equation, that do not satisfy that original equation.

Here, there simply are no real numbers satisfying [itex]sin^2(x)= 5/4[/itex] because sine is always between -1 and 1.

That's exactly the same situation as if you were trying to solve [itex]x^3+ x= 0[/itex]. [itex]x^3+ x= x(x^2+ 1)= 0[/itex] so either x= 0 or [itex]x^2= -1[/itex]. There is no real number satisfying the second equation.