Question

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

Answer 1

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?

Answer 2

\[\cos ^2\theta-\sin ^2\theta=\1-sin \theta\]
\[1-\sin ^2\theta-\sin ^2\theta=1-\sin \theta\]
\[1-2\sin ^2\theta=1-\sin \theta \]
\[2 \sin ^2 \theta-\sin \theta=0\]

Answer 3

1-sin^2 (theta)

Answer 4

okay...now what do we have to do next?

Answer 5

well... @surjithayer kinda expanded it above.... lemme show you what he did

Answer 6

\[\sin \theta \left( 2\sin \theta-1 \right)=0\]

Answer 7

Parenthesis would have been nice.

Answer 8

\(\color{blue}{\text{Originally Posted by}}\) @surjithayer
\[(\color{red}{\cos ^2\theta})-\sin ^2\theta=1-\sin \theta\]
\[(\color{red}{1-\sin ^2\theta})-\sin ^2\theta=1-\sin \theta\]
\[1-2\sin ^2\theta=1-\sin \theta \]
\[2 \sin ^2 \theta-\sin \theta=0\]
\(\color{blue}{\text{End of Quote}}\)

Answer 9

it says the final answers ought to be in radians

Answer 10

Ok? Just solve for \(\sin(\theta)\)

Answer 11

Don't use brown @jdoe0001 haha :\

Answer 12

hmmm hold the mayo... the ones should cancel out

Answer 13

hehe

Answer 14

lemme redo it some :)

Answer 15

\(\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 =)

Answer 16

much better \(\checkmark\)

Answer 17

lol...so the answers are {0 , pi , pi/6 , 5pi/6 , }...is that right?

Answer 18

well... hmm what angles does it give you a sine of 0?

Answer 19

0 and pi

Answer 20

i am not really sure so help me out

Answer 21

hmmm hold... shoot... just notice... missed somethjing, OR it was added after I read it :/

Answer 22

anyhow, @surjithayer's is correct either way :)

Answer 23

okay...i really just wanna get to the correct answer..lol...i have only two chances left

Answer 24

well... I just noticed it was = 1-sine.... thus

Answer 25

which @surjithayer has correct :)

Answer 26

okay..bit it said wrong when i put it in...the hint says use your unit circle for final answer

Answer 27

yeap... you should
lemme see if I can redo all that, with the correct equal value quick

but @surjithayer is correct, is a quadratic

Answer 28

\(\bf -sin^2(\theta)+{\color{brown}{ cos^2(\theta)}}=1-sin(\theta)
\\ \quad \\
-sin^2(\theta)+{\color{brown}{ 1-sin^2(\theta)}}=1-sin(\theta)\implies 1-2sin^2(\theta)=1-sin(\theta)
\\ \quad \\
\cancel{1-1}+sin(\theta)-2sin^2(\theta)=0\implies -2sin^2(\theta)+sin(\theta)=0
\\ \quad \\
2sin^2(\theta)-sin(\theta)=0\implies 2sin^2(\theta)-sin(\theta)+0=0\)

notice the quadratic

Answer 29

yeah

Answer 30

then just solve it like you'd any quadratic
then take the inverse sine of it

Answer 31

\[either \sin \theta=0=\sin n \pi\]
in the given interval \[\theta=\pi ,2 \pi\]

Answer 32

hmmm actually, just a quick common factor should do

Answer 33

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

Answer 34

\(\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}
}\)

Answer 35

\[or \sin \theta=\frac{ 1 }{ 2 }\]
rejected because in [pi,2 pi] sin is negative.

Answer 36

yeah that too...okay so i will put in {pi , pi/6 and 5pi/6}...i cant get this wrong again

Answer 37

omg!!...it says wrong again!!! i am done...:-((

Answer 38

only pi and 2 pi

Answer 39

its void now...but i would like to know how you got your answers from the unit circle

Answer 40

okay what about sin (theta) = 1/2

Answer 41

i say it is positive ,so not in [pi,2 pi]
it is in [0,2pi] which you don't want.

Answer 42

0

Answer 43

lol...thank you very much @surjithayer

Answer 44

correction\[\sin\ theta\]=1/2, it is in [0,pi]

Answer 45

yw