Respuesta :
Answer:
The solutions on the given interval are :
[tex]0[/tex]
[tex]\pi[/tex]
[tex]\cos^{-1}(\frac{1}{4})[/tex]
[tex]-\cos^{-1}(\frac{1}{4})+2\pi[/tex]
Step-by-step explanation:
We will need the double angle identity [tex]\sin(2x)=2\sin(x)\cos(x)[/tex].
Let's begin:
[tex]2\sin(2x)=\sin(x)[/tex]
Use double angle identity mentioned on left hand side:
[tex]2\cdot 2\sin(x)\cos(x)=\sin(x)[/tex]
Simplify a little bit on left side:
[tex]4\sin(x)\cos(x)=\sin(x)[/tex]
Subtract [tex]\sin(x)[/tex] on both sides:
[tex]4\sin(x)\cos(x)-\sin(x)=0[/tex]
Factor left hand side:
[tex]\sin(x)[4\cos(x)-1]=0[/tex]
Set both factors equal to 0 because at least of them has to be 0 in order for the equation to be true:
[tex]\sin(x)=0 \text{ or } 4\cos(x)-1=0[/tex]
The first is easy what angles [tex]\theta[/tex] are [tex]y[/tex]-coordinates on the unit circle 0. That happens at [tex]0[/tex] and [tex]\pi[/tex] on the given range of [tex]x[/tex] (this [tex]x[/tex] is not be confused with the [tex]x[/tex]-coordinate).
Now let's look at the second equation:
[tex]4\cos(x)-1=0[/tex]
Isolate [tex]\cos(x)[/tex].
Add 1 on both sides:
[tex]4 \cos(x)=1[/tex]
Divide both sides by 4:
[tex]\cos(x)=\frac{1}{4}[/tex]
This is not as easy as finding on the unit circle.
We know [tex]\arccos( )[/tex] will render us a value between [tex]0[/tex] and [tex]2\pi[/tex].
So one solution on the given interval for x is [tex]x=\cos^{-1}(\frac{1}{4})[/tex].
We know cosine function is even.
So an equivalent equation is:
[tex]\cos(-x)=\frac{1}{4}[/tex]
Apply [tex]\cos^{-1}[/tex] to both sides:
[tex]-x=\cos^{-1}(\frac{1}{4})[/tex]
Multiply both sides by -1:
[tex]x=-\cos^{-1}(\frac{1}{4})[/tex]
This going to be negative in the 4th quadrant but if we wrap around the unit circle, [tex]2\pi[/tex] , we will get an answer between [tex]0[/tex] and [tex]2\pi[/tex].
So the solutions on the given interval are :
[tex]0[/tex]
[tex]\pi[/tex]
[tex]\cos^{-1}(\frac{1}{4})[/tex]
[tex]-\cos^{-1}(\frac{1}{4})+2\pi[/tex]
