Question

find the exact solutions of the given equation in the interval [0,2pi]
cos2x+3cosx+2=0

Answer 1

\[\cos^2(x)+3\cos(x)+2=0?\]

Answer 2

it is cos(2x) +3cos(x)+2=0

Answer 3

\[(z+1)(z+2)=0\]
\[z=-1\]
\[z=-2\]
\[\cos(x)=-1\]
\[\cos(x)=-2\] second one is very unlikely

Answer 4

im sorry im not sure i follow. they have to be in the interval [0,2pi]

Answer 5

cos (2x) = 2 cos^2 x - 1

2 cos^2 x + 3 cos x + 1 = 0

Answer 6

\[\cos(2x)=\cos^2(x)-\sin^2(x)=\cos^2(x)-(1-\cos^2(x))=2\cos^2(x)-1\]

Answer 7

\[2\cos^2(x)-1+3\cos(x)+2=0\]
\[2\cos^2(x)+3\cos(x)+2-1=0\]
\[2\cos^2(x)+3\cos(x)+1=0\]
Can you factor?
\[2u^2+3u+1=0\]
(note the relationship is that u=cos(x))

Answer 8

\[2u^2+2u+1u+1=0\]
since 2u+1u=3u
now we factor by grouping
\[2u(u+1)+1(u+1)=0\]
\[(u+1)(2u+1)=0\]

Answer 9

=>\[u=-1 \text{ or } u=\frac{-1}{2}\]

Answer 10

but remember u =cos(x)

Answer 11

\[\cos(x)=-1 \text{ or } \cos(x)=\frac{-1}{2}\]

Answer 12

yes i got it now! thank you :)))