Question

4 sin^2 x + 6 sin x + 2 = 0
Find all solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)

Answer 1

Someone please help me!

Answer 2

\[4\sin ^2x+6\sin x+2=0,\]
divide by 2
\[2\sin ^2x+3\sin x+1=0\]
make factors and find value of sinx
(remember \[\left| \sin x \right| \leq 1\]

Answer 3

and finally find the values of x in [0,2 pi]

Answer 4

I don't understand how to find the values of x [0,2 pi]

Answer 5

[0,2pi) *

Answer 6

So how many solutions in all are there? It says to enter answers as a comma-separated list.

Answer 7

I just do not understand how it relates to the unit circle

Answer 8

3mar that solution set is not correct D:
Hmm I'm not sure where you came up with those...

Answer 9

\[2 \sin ^2x+2\sin x+\sin x+1=0\]
\[2\sin x \left( \sin x+1 \right)+1(\sin x+1)=0\]
(sinx+1)(2 sin x+1)=0
sin x=-1=sin3pi/2
\[x=\frac{ 3 \pi }{ 2 }\]
\[\sin x=-\frac{ 1 }{ 2 }=-\sin \frac{ \pi }{ 6 }=\sin \left( \pi+\frac{ \pi }{ 6 } \right),\sin \left( 2 \pi-\frac{ \pi }{ 6 } \right)\]
\[x=\frac{ 7 \pi }{ 6 },\frac{ 11 \pi }{ 6 }\]