Question

Please help me understand this trig!
2sin^2(x) + 3sin(x) + 1 = 0
For some reason my teacher decided not to teach this lesson and I am completely lost. :(

Answer 1

Can you solve 2u^2+3u+1=0

Answer 2

call sinx=a then
2a^2+3a+1=0
(2a+1).(a+1)=0
a=-1/2 and a=-1
sin(x)=-1/2 and sin(x)=-1
for -1/2 x=210 and 330
for -1 x=270

Answer 3

let \(\sin(x) = y\) then you have
\(2y^2+3y+1=0\\2y^2+2y+y+1=0\\2y(y+1)+(y+1)=0\\(2y+1)(y+1)=0\\y=\frac{-1}{2},y=-1\\\implies \sin(x)=\frac{-1}{2}, \sin(x) = -1\)

Answer 4

I know how to do all this work but for my final answer is it \[-\pi/6\]and \[-\pi/2\]

Answer 5

there are infinite answers

Answer 6

do I add 2pi(n) or pi(n)? or none?

Answer 7

\(\sin(x) = -1 \implies x = \frac{3\pi}{2}\pm2\pi k \) where \(k\in \mathbb{Z}\)

Answer 8

between interval [0, 2pi)

Answer 9

ahh

Answer 10

sorry \(\sin(x) = -1 \implies x = \frac{3\pi}{2}\)
\(\sin(x) = \frac{-1}{2}\implies x = \frac{7\pi}{6}, \frac{11\pi}{6}\)

Answer 11

okay thank you - so i don't add 2pi(n) of pi(n)? If not for this problem, is there a problem where you would?

Answer 12

@zzr0ck3r so i don't add 2pi(n) of pi(n)? If not for this problem, is there a problem where you would?