Ask
your own question, for FREE!
Mathematics
14 Online
Solve on the interval [0,2pi): 2sin^2(x)+3sinx+1=0 A. x=pi,x=2pi/3,x=5pi/3 B. x=2pi,x=pi/3 C. x=3pi/2,x=7pi/6,x=11pi/6 D. x=2pi,x=pi/4,x=5pi/4
Still Need Help?
Join the QuestionCove community and study together with friends!
factor it and set each factor equal to 0
[x = -(1/6)*Pi], [x = -(1/2)*Pi]
I hate it when someone just spouts out the answers.
Can't find your answer?
Make a FREE account and ask your own questions, OR help others and earn volunteer hours!
Join our real-time social learning platform and learn together with your friends!
Join our real-time social learning platform and learn together with your friends!
Latest Questions
Ferrari:
I Have Returned to Writing Music (by me) so if you don't know, I used to write songs on here and post them, some of them were lowkey corny though.
kaelynw:
tried a lil smt, the arm is off but i like the other stuff
Countless7Echos:
whaa this looks so silly, nothing is funnier then looking at how I draw my hands
Countless7Echos:
I don't know just no sketch doodle day :p finished a video already so I'm pretty
1 minute ago
34 Replies
2 Medals
15 minutes ago
10 Replies
2 Medals
8 hours ago
8 Replies
0 Medals
1 day ago
9 Replies
2 Medals