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Mathematics
OpenStudy (anonymous):

Find the solution to sqrt(2)sinx+1=0

OpenStudy (anonymous):

do u mean \[\sqrt{2} (x) + 1 =0\]

OpenStudy (anonymous):

oh, sorrry. it is like that, but with a sin before the x

OpenStudy (anonymous):

substract 1 from both sides, divide both sides by root 2. and then sin-1(result)

jimthompson5910 (jim_thompson5910):

sqrt(2)sinx+1=0 sqrt(2)sinx=-1 sin(x) = -1/(sqrt(2)) sin(x) = -sqrt(2)/2 Then use the unit circle to determine which values of x (or which angles) make this equation true.

OpenStudy (anonymous):

so it is 3pi/4?

jimthompson5910 (jim_thompson5910):

no because sin(3pi/4) = sqrt(2)/2

OpenStudy (anonymous):

sorry, didn't see the -

jimthompson5910 (jim_thompson5910):

no, sin(pi/4) = sqrt(2)/2

jimthompson5910 (jim_thompson5910):

It turns out that 5pi/4 works though since sin(5pi/4) = -sqrt(2)/2 Remember to look below the x axis on the unit circle (since sine is negative)

OpenStudy (anonymous):

would 7pi/4 work as well?

jimthompson5910 (jim_thompson5910):

yes it would

OpenStudy (anonymous):

Thank You!!!

jimthompson5910 (jim_thompson5910):

Are you solutions restricted to a certain interval?

jimthompson5910 (jim_thompson5910):

like do you have to keep the solutions within [0,2pi) or something?

OpenStudy (anonymous):

I have 7pi/4 as a choice, but i didn't have 5pi/4

jimthompson5910 (jim_thompson5910):

ah so it's multiple choice, nvm that thought then

OpenStudy (anonymous):

Oh ok... thanks for thinking of it anyway though :)

jimthompson5910 (jim_thompson5910):

you're welcome

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