# `(sin^2theta)/(1+cos theta)=1` What are the values of `theta` in the interval below that satisfy the equation? `0<=theta<=360`

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### 2 Answers

Solve `(sin^2theta)/(1+costheta)=1` for `0<=theta<=360^@` :

`(sin^2theta)/(1+costheta)=1`

`sin^2theta=1+costheta` Use the Pythagorean relationship

`1-cos^2theta=1+costheta`

`cos^2theta+costheta=0`

`costheta(costheta+1)=0`

`costheta=0==>theta=90^@,270^@`

`costheta=-1==>theta=180^@` but this is extraneous -- the fraction is undefined at `theta=180^@`

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The solutions are `theta=90^@,270^@`

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The graph of `y=(sin^2theta)/(1+costheta)` and y=1:

**The graph is in radians so `theta=pi/2,(3pi)/2` **

*** Sorry about the costhheta -- it will not let me correct them. Hover over them and you will see that they are typed in correctly.

### User Comments

It is better to use radians instead of degrees for such a question.

The domain is 0 to Pi inclusive.

Sin^2x=1+cosx

Substituting (Cosx)^2+(Sinx)^2 for 1.

Sin^2x=Cos^2x+Sin^2x+Cosx

Solving for zero, and canceling like terms.

0=Cos^2x+Cosx

Factoring out the cosx's

0=Cosx(Cosx+1)

Where each seperate factor of the formula have to equal zero to get our desired solutions.

They turn out to be: Pi/2 and Pi