Respuesta :
Answer:
[tex]\theta \in \{0,\pi,\frac{7\pi}{6},\frac{11\pi}{6}\}[/tex]
Step-by-step explanation:
[tex]\sin(\theta)+1=\cos(2\theta)[/tex]
Applying double angle identity:
[tex]\cos(2\theta)=1-2\sin^2(\theta)[/tex]
Doing so would give:
[tex]\sin(\theta)+1=1-2\sin^2(\theta)[/tex]
We need to get everything to one side so we have 0 on one side.
Subtract 1 on both sides:
[tex]\sin(\theta)=-2\sin^2(\theta)[/tex]
Add [tex]2\sin^2(theta)[/tex] on both sides:
[tex]\sin(\theta)+2\sin^2(\theta)=0[/tex]
Let's factor the left-hand side.
The two terms on the left-hand side have a common factor of [tex]\sin(\theta)[/tex].
[tex]\sin(\theta)[1+2\sin(\theta)]=0[/tex].
This implies we have:
[tex]\sin(\theta)=0 \text{ or } 1+2\sin(\theta)=0[/tex].
We need to solve both equations.
You are asking they be solved in the interval [tex][0,2\pi)[/tex].
[tex]\sin(\theta)=0[/tex]
This means look at your unit circle and find when you have your y-coordinates is 0.
You this at 0 and [tex]\pi[/tex]. (I didn't include [tex]2\pi[/tex] because you don't have a equal sign at the endpoint of [tex]2\pi[/tex].
Now let's solve [tex]1+2\sin(\theta)=0[/tex]
Subtract 1 on both sides:
[tex]2\sin(\theta)=-1[/tex]
Divide both sides by 2:
[tex]\sin(\theta)=\frac{-1}{2}[/tex]
Now we are going to go and look for when the y-coordinates are -1/2.
This happens at [tex]\frac{7\pi}{6}[/tex] and [tex]\frac{11\pi}{6}[/tex].
The solution set given the restrictions is
[tex]\theta \in \{0,\pi,\frac{7\pi}{6},\frac{11\pi}{6}\}[/tex]
