Question

sqrt2 sin^2-sin=0
solve the equation

Answer 1

You mean:

\[\sqrt{2\sin^2(x) - \sin(x)} = 0\]

Answer 2

@ashleedean12 want to reply or not??

Answer 3

\[\sqrt{2}\sin ^{2}\theta-\sin \theta=0\]

Answer 4

@waterineyes

Answer 5

Where were you??

Answer 6

working on a practice test

Answer 7

What you can do here, just take one sin(theta) out of the equation, I mean to factor out one sin(theta)..

Can you do that??

Answer 8

i have no idea how to do that

Answer 9

I am just giving you an example:

\[\cos^2(\theta) - \cos(\theta) = 2 \implies \color{red}{\cos(\theta) (\cos(\theta) - 1 ) = 2}\]

Answer 10

Like this taking one common out of the equation..

Same you can apply there in your question.

Answer 11

@yummydum will explain you better than me...

Answer 12

i havent done this in a while but i think this is how:

\[\sqrt{2}\sin^2\theta-\sin\theta=0\]\[\sin\theta(\sqrt{2}\sin\theta-1)=0\]\[\sin\theta=0~~~~~~~~~~~~~~~~~~~~\sqrt{2}\sin\theta-1=0\]\[\theta=\sin^{-1}(0)~~~~~~~~~~~~~~\sqrt{2}\sin\theta=1\]\[\theta=0,\pi+\pi n~~~~~~~~~~~~~~\sin\theta={1\over\sqrt{2}}\]\[~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\theta=\sin^{-1}\left(1\over\sqrt{2}\right)\]\[~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\theta={\pi\over4}+2\pi n, {3\pi\over4}+\pi n\]

and so, \(\theta=0,~\pi+\pi n,~{\pi\over4}+2\pi n,~{3\pi\over4}+\pi n\)

Answer 13

so sqrt2 -1sin -sin=0

Answer 14

no, look at what i did

Answer 15

that confuses me

Answer 16

so would the answer be 0,pi, pi/4, 3pi/4

Answer 17

Ha ha ha ha... @yummydum

Answer 18

Don't go with the answers @ashleedean12

Just understand the solution that @yummydum has given above..

Answer 19

you factor out a \(\sin\theta\) and now you have two factors, \(\sin\theta\) and \(\sqrt{2}\sin\theta-1\), both of which are equal to \(0\)
separately set the factors equal to \(0\) and solve.....you have to know the unit circle to know what \(\theta\) is at \(0\) and \({1\over\sqrt{2}}\)

Answer 20

\[\large \color{red} {and~~so,~~\theta=0,~\pi+\pi n,~{\pi\over4}+2\pi n,~{3\pi\over4}+\pi n}\]