Question

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

Answer 1

Answer:

$\theta \in \{0,\pi,\frac{7\pi}{6},\frac{11\pi}{6}\}$

Step-by-step explanation:

$\sin(\theta)+1=\cos(2\theta)$

Applying double angle identity:

$\cos(2\theta)=1-2\sin^2(\theta)$

Doing so would give:

$\sin(\theta)+1=1-2\sin^2(\theta)$

We need to get everything to one side so we have 0 on one side.

Subtract 1 on both sides:

$\sin(\theta)=-2\sin^2(\theta)$

Add $2\sin^2(theta)$ on both sides:

$\sin(\theta)+2\sin^2(\theta)=0$

Let's factor the left-hand side.

The two terms on the left-hand side have a common factor of $\sin(\theta)$.

$\sin(\theta)[1+2\sin(\theta)]=0$.

This implies we have:

$\sin(\theta)=0 \text{ or } 1+2\sin(\theta)=0$.

We need to solve both equations.

You are asking they be solved in the interval $[0,2\pi)$.

$\sin(\theta)=0$

This means look at your unit circle and find when you have your y-coordinates is 0.

You this at 0 and $\pi$. (I didn't include $2\pi$ because you don't have a equal sign at the endpoint of $2\pi$.

Now let's solve $1+2\sin(\theta)=0$

Subtract 1 on both sides:

$2\sin(\theta)=-1$

Divide both sides by 2:

$\sin(\theta)=\frac{-1}{2}$

Now we are going to go and look for when the y-coordinates are -1/2.

This happens at $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$.

The solution set given the restrictions is

$\theta \in \{0,\pi,\frac{7\pi}{6},\frac{11\pi}{6}\}$