Question

Finals this afternoon: Could someone explain this to me?
Solve the equation on the interval [-2π,0)
2(sin^2 x)+3sinx+1=0

Answer 1

let \[\sin x \] be \[a\]
\[2a ^{2}+3a+1=0\]
\[(2a+1)(a+1)=0\]
\[a=-1/2\] and
\[a=-1\]so \[\sin x=-1/2\] and \[\sin x=-1\]

Answer 2

how would i solve sinx=-1/2 and sinx=-1, would i use the unit circle or is there another way to go about solving it?

Answer 3

you can also use a calculator or special angles for 30-60-90
calculator method
\[x=\sin^{-1} (-1/2)=-30=\pi/6\]
\[x=\sin^{-1} (-1)=-90=\pi/2\]

Answer 4

so for your general solution you have
\[x=-\pi/6+2\pi.k,\pi \in Z\]
and

\[x=-\pi/2+2\pi.k \in Z \]
\[x=-2\pi,-6/5\pi,-1/2\pi,9/4\]

Answer 5

okay,,that makes sense, thank you so much for helping me,,,

Answer 6

my pleasure some steps are omitted in getting x