Question

For 0 ≤ ϴ < 2π, how many solutions are there to tan(StartFraction theta Over 2 EndFraction) = sin(ϴ)? Note: Do not include values that are undefined for tan or sin(ϴ).

Answer 1

its D. 3

Explanation:

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Answer 2

3 solutions:

`\theta={0, (\pi)/(2), (3\pi)/(2)}`

Explanation:

So, first of all, we need to figure the angles that cannot be included in our answers out. The only function in the equation that isn't defined for some angles is
`tan((\theta)/(2))` so let's focus on that part of the equation first.

We know that:

`tan((\theta)/(2))=(sin((\theta)/(2)))/(cos((\theta)/(2)))`

therefore:

`cos((\theta)/(2))\\eq0`

so we need to find the angles that will make the cos function equal to zero. So we get:

`cos((\theta)/(2))=0`

`(\theta)/(2)=cos^(-1)(0)`

`(\theta)/(2)=(\pi)/(2)+\pi n`

or

`\theta=\pi+2\pi n`

we can now start plugging values in for n:

`\theta=\pi+2\pi (0)=\pi`

if we plugged any value greater than 0, we would end up with an angle that is greater than
`2\pi` so, that's the only angle we cannot include in our answer set, so:

`\theta\\eq \pi`

having said this, we can now start solving the equation:

`tan((\theta)/(2))=sin(\theta)`

we can start solving this equation by using the half angle formula, such a formula tells us the following:

`tan((\theta)/(2))=(1-cos(\theta))/(sin(\theta))`

so we can substitute it into our equation:

`(1-cos(\theta))/(sin(\theta))=sin(\theta)`

we can now multiply both sides of the equation by
`sin(\theta)`

so we get:

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

we can use the pythagorean identity to rewrite
`sin^(2)(\theta)` in terms of cos:

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

so we get:

`1-cos(\theta)=1-cos^(2)(\theta)`

we can subtract a 1 from both sides of the equation so we end up with:

`-cos(\theta)=-cos^(2)(\theta)`

and we can now add
`cos^(2)(\theta)`

to both sides of the equation so we get:

`cos^(2)(\theta)-cos(\theta)=0`

and we can solve this equation by factoring. We can factor
`cos(\theta)` to get:

`cos(\theta)(cos(\theta)-1)=0`

and we can use the zero product property to solve this, so we get two equations:

Equation 1:

`cos(\theta)=0`

`\theta=cos^(-1)(0)`

`\theta={(\pi)/(2), (3\pi)/(2)}`

Equation 2:

`cos(\theta)-1=0`

we add a 1 to both sides of the equation so we get:

`cos(\theta)=1`

`\theta=cos^(-1)(1)`

`\theta=0`

so we end up with three answers to this equation:

`\theta={0, (\pi)/(2), (3\pi)/(2)}`