Question

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

`\theta \in \{0,\pi,(7\pi)/(6),(11\pi)/(6)\}`

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)=(-1)/(2)`

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

This happens at
`(7\pi)/(6)` and
`(11\pi)/(6)`.

The solution set given the restrictions is

`\theta \in \{0,\pi,(7\pi)/(6),(11\pi)/(6)\}`