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Mathematics 16 Online
OpenStudy (dtan5457):

Find the general solution.

OpenStudy (dtan5457):

\[sinx+\sqrt{2}=-sinx\]

OpenStudy (dtan5457):

I got sinx=-sqrt2/2

OpenStudy (dtan5457):

3rd and 4th quadrants?

OpenStudy (dtan5457):

225 degrees+2npi 315 degrees+2npi ?

OpenStudy (dtan5457):

@mathmath333

OpenStudy (idku):

\[\sin(x)+\sqrt{2}=-\sin(x)\]\[\sqrt{2}=-2\sin(x)\]\[-\sqrt{2}/2=\sin(x)\]

OpenStudy (idku):

|dw:1427075714786:dw|

OpenStudy (dtan5457):

Yes..

OpenStudy (dtan5457):

lol

OpenStudy (idku):

so, your solution is pretty much that, but in a different quadrant.... good luck !

OpenStudy (dtan5457):

5pi/3+2npi and converting 315 degrees rn..

OpenStudy (dtan5457):

7pi/4+2npi

OpenStudy (dtan5457):

correct?

OpenStudy (idku):

I am getting \[2 \pi n-\frac{\pi}{4}\]\[2 \pi n+\frac{5\pi}{4}\]

OpenStudy (dtan5457):

Oh yeah, I converted 225 degrees incorrectly

OpenStudy (dtan5457):

but what about 7pi/4+2npi?

OpenStudy (dtan5457):

i mean considering if i couldn't use -45 degrees as an answer

OpenStudy (idku):

all these pi's.... I will check wolfram real quick.

OpenStudy (dtan5457):

alright thanks

OpenStudy (idku):

he is talking 1/4 th out

OpenStudy (dtan5457):

i think im right

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