Question

find all solutions to the equation

sqt(2) sinx+1=0

Answer 1

x=n(pi) +[(-1)^n+1]*pi/4

Answer 2

Now put integral values of n in that equation to get all the answers

Answer 3

\[\sin x=\frac{ -1 }{ \sqrt{2} }=-\sin \frac{ \pi }{4 }=\sin \frac{ -\pi }{4 },x=\frac{- \pi }{4 }\]
\[for~[0,2\pi],\sin x=\sin \frac{ -\pi }{ 4 }=\sin \left( \pi+\frac{ \pi }{ 4 } \right),\sin \left( 2\pi-\frac{ \pi }{ 4} \right)\]
\[x=\frac{ 5\pi }{ 4 },\frac{ 7\pi }{ 4 }\]

Answer 4

sin x = -1/sqrt(2) will be true for 180+45o and 270+45o

and then for all angles 360 or multiples of 360 degrees greater or less than these.

Answer 5

Since sine is a periodic function with domain -infinity to infinity it will have infinite answers. Use the above equation to get the ones in the given domain

Answer 6

I wasn't given a domain, they are asking for all the solutions...I am so confused

Answer 7

What are the options?

Answer 8

there are none

Answer 9

Then there's nothing you can do except give the ones in the usual domains say -2pi to 2pi

Answer 10

sqrt{2}sin (x)+1=0 is all the give me

Answer 11

Substitute n=-3,-2,-1,0,1,2,3 and write them down. Because really there are infinite answers if no domain is given

Answer 12

fantastic

Answer 13

They probably want 180+45=224 and 270+45=315 degrees.