Question

How do you solve 2sin^2x+3sinx+1=0? for greater than or equal to 0 or less than 360?

Answer 1

When I calculated it, I got a bunch of answers on wolfram, but I don't know how to get it and or if these are right.

Answer 2

\[2\sin ^{2}x+3\sin x+1=0\]

Answer 3

Can you help when you get a chance. It is the same type of question, sorry. @satellite73

Answer 4

So far, I put 1 on the other side and took out sinx, but I don't know if this is right.

Answer 5

Will be your fan if you help, I really need to finish this. Sorry to sound desperate, but I am.

Answer 6

\[2\sin ^2x+2\sin x+\sin x+1=0\]
\[2\sin x \left( \sin x+1 \right)+1\left( \sin x+1 \right)=0\]
\[\left( \sin x+1 \right)\left( 2\sin x+1 \right)=0\]
\[Either~\sin x=-1=-\sin 90=\sin \left( 180+90 \right)=\sin 270,x=270 degree\]
\[or ~\sin x=-\frac{ 1 }{2 }=-\sin 30=\sin \left( 180+30 \right)=\sin 210,x=210~degree\]

Answer 7

Ok, that helps a lot. Thanks!