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

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myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\cos^2(x)+3\cos(x)+2=0?\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is cos(2x) +3cos(x)+2=0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(z+1)(z+2)=0\] \[z=1\] \[z=2\] \[\cos(x)=1\] \[\cos(x)=2\] second one is very unlikely

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im sorry im not sure i follow. they have to be in the interval [0,2pi]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cos (2x) = 2 cos^2 x  1 2 cos^2 x + 3 cos x + 1 = 0

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\cos(2x)=\cos^2(x)\sin^2(x)=\cos^2(x)(1\cos^2(x))=2\cos^2(x)1\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[2\cos^2(x)1+3\cos(x)+2=0\] \[2\cos^2(x)+3\cos(x)+21=0\] \[2\cos^2(x)+3\cos(x)+1=0\] Can you factor? \[2u^2+3u+1=0\] (note the relationship is that u=cos(x))

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[2u^2+2u+1u+1=0\] since 2u+1u=3u now we factor by grouping \[2u(u+1)+1(u+1)=0\] \[(u+1)(2u+1)=0\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1=>\[u=1 \text{ or } u=\frac{1}{2}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1but remember u =cos(x)

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\cos(x)=1 \text{ or } \cos(x)=\frac{1}{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes i got it now! thank you :)))
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