Answer:
[tex]\{0,\pi , \frac{7\pi}{6},\frac{11\pi}{6}\}[/tex]
Step-by-step explanation:
We are going to have to use a double angle identity.
[tex]\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)[/tex]
I'm going to write this in terms of sine since the left hand side of the equation is also in terms of sine.
So applying the Pythagorean Identity [tex]\sin^2(\theta)+\cos^2(\theta)=1[/tex] to the identity above gives you:
[tex]\cos(2\theta)=(1-\sin^2(\theta))-\sin^2(\theta)[/tex]
[tex]\cos(2\theta)=1-2\sin^2(\theta)[/tex]
So the equation you have becomes:
[tex]\sin(\theta)+1=1-2\sin^2(\theta)[/tex]
Get one side to be 0.
I'm going to move everything on right side to left side.
When you move a term over from one side to another, you just need to change the sign in front it. so if is plus 1 on the right it is minus 1 on the left.
If it is minus [tex]2\sin^2(\theta)[/tex] then it is plus [tex]2\sin^2(\theta)[/tex] on the other side.
[tex]2\sin^2(\theta)+\sin(\theta)+1-1=0[/tex]
Combine the like terms (1-1):
[tex]2\sin^2(\theta)+\sin(\theta)+0=0[/tex]
[tex]2\sin^2(\theta)+\sin(\theta)=0[/tex]
Factor out the common factor on the left which is sin( )[/tex]
[tex]\sin(\theta)[2\sin(\theta)+1]=0[/tex]
If you have a product is 0 then one of the factors must be zero (both could be 0 also).
[tex]\sin(\theta)=0[/tex] or [tex]2\sin(\theta)+1=0[/tex]
I'm going to solve the first equation first.
[tex]\sin(\theta)=0[/tex] when the y-coordinate on the unit circle is 0.
This happens at [tex]0 \text{ or } \pi[/tex] is the given interval you provided.
Let's solve the other equation:
[tex]2\sin(\theta)+1=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]
So you are looking for when the y-coordinate is -1/2 on the unit circle.
This happens at [tex]\frac{7\pi}{6} \text{ or } \frac{11\pi}{6}[/tex]
So the solution set is:
[tex]\{0,\pi , \frac{7\pi}{6},\frac{11\pi}{6}\}[/tex]
