Solve within interval [pi,2pi] -sin(squared)theta+cos(squared)theta=1-sin(theta)

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Solve within interval [pi,2pi] -sin(squared)theta+cos(squared)theta=1-sin(theta)

Mathematics
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well.... 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?
\[\cos ^2\theta-\sin ^2\theta=\1-sin \theta\] \[1-\sin ^2\theta-\sin ^2\theta=1-\sin \theta\] \[1-2\sin ^2\theta=1-\sin \theta \] \[2 \sin ^2 \theta-\sin \theta=0\]
1-sin^2 (theta)

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okay...now what do we have to do next?
well... @surjithayer kinda expanded it above.... lemme show you what he did
\[\sin \theta \left( 2\sin \theta-1 \right)=0\]
Parenthesis would have been nice.
\(\color{blue}{\text{Originally Posted by}}\) @surjithayer \[(\color{red}{\cos ^2\theta})-\sin ^2\theta=1-\sin \theta\] \[(\color{red}{1-\sin ^2\theta})-\sin ^2\theta=1-\sin \theta\] \[1-2\sin ^2\theta=1-\sin \theta \] \[2 \sin ^2 \theta-\sin \theta=0\] \(\color{blue}{\text{End of Quote}}\)
it says the final answers ought to be in radians
Ok? Just solve for \(\sin(\theta)\)
Don't use brown @jdoe0001 haha :\
hmmm hold the mayo... the ones should cancel out
hehe
lemme redo it some :)
\(\bf -sin^2(\theta)+cos^2(\theta)=1 \\ \quad \\ sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)={\color{blue}{ 1-sin^2(\theta) }}\qquad thus \\ \quad \\ -sin^2(\theta)+cos^2(\theta)=1\implies -sin^2(\theta)+[{\color{blue}{ 1-sin^2(\theta) }}]=1 \\ \quad \\ 1-sin^2(\theta)-sin^2(\theta)=1\implies 1-2sin^2(\theta)=1 \\ \quad \\ 0=2sin^2(\theta)\implies 0=sin^2(\theta)\implies 0=sin(\theta) \\ \quad \\ sin^{-1}(0)=sin^{-1}\left[ sin(\theta) \right]\implies sin^{-1}(0)=\theta\) blue then =)
much better \(\checkmark\)
lol...so the answers are {0 , pi , pi/6 , 5pi/6 , }...is that right?
well... hmm what angles does it give you a sine of 0?
0 and pi
i am not really sure so help me out
hmmm hold... shoot... just notice... missed somethjing, OR it was added after I read it :/
anyhow, @surjithayer's is correct either way :)
okay...i really just wanna get to the correct answer..lol...i have only two chances left
well... I just noticed it was = 1-sine.... thus
which @surjithayer has correct :)
okay..bit it said wrong when i put it in...the hint says use your unit circle for final answer
yeap... you should lemme see if I can redo all that, with the correct equal value quick but @surjithayer is correct, is a quadratic
\(\bf -sin^2(\theta)+{\color{brown}{ cos^2(\theta)}}=1-sin(\theta) \\ \quad \\ -sin^2(\theta)+{\color{brown}{ 1-sin^2(\theta)}}=1-sin(\theta)\implies 1-2sin^2(\theta)=1-sin(\theta) \\ \quad \\ \cancel{1-1}+sin(\theta)-2sin^2(\theta)=0\implies -2sin^2(\theta)+sin(\theta)=0 \\ \quad \\ 2sin^2(\theta)-sin(\theta)=0\implies 2sin^2(\theta)-sin(\theta)+0=0\) notice the quadratic
yeah
then just solve it like you'd any quadratic then take the inverse sine of it
\[either \sin \theta=0=\sin n \pi\] in the given interval \[\theta=\pi ,2 \pi\]
hmmm actually, just a quick common factor should do
i 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
\(\large { 2sin^2(\theta)-sin(\theta)=0\implies {\color{brown}{ sin(\theta)}}[2sin(\theta)-1]=0 \\ \quad \\ \begin{cases} sin(\theta)=0\to &\theta=sin^{-1}(0)\\ \quad \\ 2sin(\theta)-1=0\to sin(\theta)=\frac{1}{2}\to &\theta=sin^{-1}\left( \frac{1}{2} \right) \end{cases} }\)
\[or \sin \theta=\frac{ 1 }{ 2 }\] rejected because in [pi,2 pi] sin is negative.
yeah that too...okay so i will put in {pi , pi/6 and 5pi/6}...i cant get this wrong again
omg!!...it says wrong again!!! i am done...:-((
only pi and 2 pi
its void now...but i would like to know how you got your answers from the unit circle
|dw:1444534632816:dw|
okay what about sin (theta) = 1/2
i say it is positive ,so not in [pi,2 pi] it is in [0,2pi] which you don't want.
0
lol...thank you very much @surjithayer
correction\[\sin\ theta\]=1/2, it is in [0,pi]
yw

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