Solve sin θ+1= cos2θ on the interval 0≤ θ<2 pi. Show work

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]