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Mathematics
OpenStudy (anonymous):

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

myininaya (myininaya):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

cos (2x) = 2 cos^2 x - 1 2 cos^2 x + 3 cos x + 1 = 0

myininaya (myininaya):

$\cos(2x)=\cos^2(x)-\sin^2(x)=\cos^2(x)-(1-\cos^2(x))=2\cos^2(x)-1$

myininaya (myininaya):

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

myininaya (myininaya):

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

myininaya (myininaya):

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

myininaya (myininaya):

but remember u =cos(x)

myininaya (myininaya):

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

OpenStudy (anonymous):

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

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