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Jhulian
 one year ago
Solve within interval [pi,2pi]
sin(squared)theta+cos(squared)theta=1sin(theta)
Jhulian
 one year ago
Solve within interval [pi,2pi] sin(squared)theta+cos(squared)theta=1sin(theta)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well.... recall that \(\bf sin^2(\theta)+cos^2(\theta)=1\) so... if you solve that for say... \(\bf cos^2\) what does that give you?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\cos ^2\theta\sin ^2\theta=\1sin \theta\] \[1\sin ^2\theta\sin ^2\theta=1\sin \theta\] \[12\sin ^2\theta=1\sin \theta \] \[2 \sin ^2 \theta\sin \theta=0\]

Jhulian
 one year ago
Best ResponseYou've already chosen the best response.0okay...now what do we have to do next?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well... @surjithayer kinda expanded it above.... lemme show you what he did

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sin \theta \left( 2\sin \theta1 \right)=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Parenthesis would have been nice.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{blue}{\text{Originally Posted by}}\) @surjithayer \[(\color{red}{\cos ^2\theta})\sin ^2\theta=1\sin \theta\] \[(\color{red}{1\sin ^2\theta})\sin ^2\theta=1\sin \theta\] \[12\sin ^2\theta=1\sin \theta \] \[2 \sin ^2 \theta\sin \theta=0\] \(\color{blue}{\text{End of Quote}}\)

Jhulian
 one year ago
Best ResponseYou've already chosen the best response.0it says the final answers ought to be in radians

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok? Just solve for \(\sin(\theta)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Don't use brown @jdoe0001 haha :\

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm hold the mayo... the ones should cancel out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lemme redo it some :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf sin^2(\theta)+cos^2(\theta)=1 \\ \quad \\ sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)={\color{blue}{ 1sin^2(\theta) }}\qquad thus \\ \quad \\ sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)+[{\color{blue}{ 1sin^2(\theta) }}]=1 \\ \quad \\ 1sin^2(\theta)sin^2(\theta)=1\implies 12sin^2(\theta)=1 \\ \quad \\ 0=2sin^2(\theta)\implies 0=sin^2(\theta)\implies 0=sin(\theta) \\ \quad \\ sin^{1}(0)=sin^{1}\left[ sin(\theta) \right]\implies sin^{1}(0)=\theta\) blue then =)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0much better \(\checkmark\)

Jhulian
 one year ago
Best ResponseYou've already chosen the best response.0lol...so the answers are {0 , pi , pi/6 , 5pi/6 , }...is that right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well... hmm what angles does it give you a sine of 0?

Jhulian
 one year ago
Best ResponseYou've already chosen the best response.0i am not really sure so help me out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm hold... shoot... just notice... missed somethjing, OR it was added after I read it :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0anyhow, @surjithayer's is correct either way :)

Jhulian
 one year ago
Best ResponseYou've already chosen the best response.0okay...i really just wanna get to the correct answer..lol...i have only two chances left

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well... I just noticed it was = 1sine.... thus

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which @surjithayer has correct :)

Jhulian
 one year ago
Best ResponseYou've already chosen the best response.0okay..bit it said wrong when i put it in...the hint says use your unit circle for final answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeap... you should lemme see if I can redo all that, with the correct equal value quick but @surjithayer is correct, is a quadratic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf sin^2(\theta)+{\color{brown}{ cos^2(\theta)}}=1sin(\theta) \\ \quad \\ sin^2(\theta)+{\color{brown}{ 1sin^2(\theta)}}=1sin(\theta)\implies 12sin^2(\theta)=1sin(\theta) \\ \quad \\ \cancel{11}+sin(\theta)2sin^2(\theta)=0\implies 2sin^2(\theta)+sin(\theta)=0 \\ \quad \\ 2sin^2(\theta)sin(\theta)=0\implies 2sin^2(\theta)sin(\theta)+0=0\) notice the quadratic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then just solve it like you'd any quadratic then take the inverse sine of it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[either \sin \theta=0=\sin n \pi\] in the given interval \[\theta=\pi ,2 \pi\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm actually, just a quick common factor should do

Jhulian
 one year ago
Best ResponseYou've already chosen the best response.0i get two equations sin Theta = 0...and sin Theta = 1/2 when I used my unit circle ti find the final answers it says wrong...lol...This is my last chance

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\large { 2sin^2(\theta)sin(\theta)=0\implies {\color{brown}{ sin(\theta)}}[2sin(\theta)1]=0 \\ \quad \\ \begin{cases} sin(\theta)=0\to &\theta=sin^{1}(0)\\ \quad \\ 2sin(\theta)1=0\to sin(\theta)=\frac{1}{2}\to &\theta=sin^{1}\left( \frac{1}{2} \right) \end{cases} }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[or \sin \theta=\frac{ 1 }{ 2 }\] rejected because in [pi,2 pi] sin is negative.

Jhulian
 one year ago
Best ResponseYou've already chosen the best response.0yeah that too...okay so i will put in {pi , pi/6 and 5pi/6}...i cant get this wrong again

Jhulian
 one year ago
Best ResponseYou've already chosen the best response.0omg!!...it says wrong again!!! i am done...:((

Jhulian
 one year ago
Best ResponseYou've already chosen the best response.0its void now...but i would like to know how you got your answers from the unit circle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444534632816:dw

Jhulian
 one year ago
Best ResponseYou've already chosen the best response.0okay what about sin (theta) = 1/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i say it is positive ,so not in [pi,2 pi] it is in [0,2pi] which you don't want.

Jhulian
 one year ago
Best ResponseYou've already chosen the best response.0lol...thank you very much @surjithayer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0correction\[\sin\ theta\]=1/2, it is in [0,pi]
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