`(sin^2theta)/(1+cos theta)=1` What are the values of `theta` in the interval below that satisfy the equation? `0<=theta<=360`
2 Answers
embizze | High School Teacher | (Level 2) Educator Emeritus
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
chainrule | (Level 1) eNoter
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