DominiRican1013
  • DominiRican1013
Solve on the interval [0,2π): 2cos2x+3cosx+1=0
Mathematics
schrodinger
  • schrodinger
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Michele_Laino
  • Michele_Laino
here we have to make this substitution: \[\cos \left( {2x} \right) = 2{\left( {\cos x} \right)^2} - 1\]
Michele_Laino
  • Michele_Laino
so your equation, can be rewritten as follows: \[2\left( {2{{\left( {\cos x} \right)}^2} - 1} \right) + 3\cos x + 1 = 0\]
Michele_Laino
  • Michele_Laino
please simplify

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Michele_Laino
  • Michele_Laino
what equation do you get?
DominiRican1013
  • DominiRican1013
I don't know I'm trying to work it out
Michele_Laino
  • Michele_Laino
hint: \[4{\left( {\cos x} \right)^2} - 2 + 3\cos x + 1 = 0\]
DominiRican1013
  • DominiRican1013
I'll be right back.
DominiRican1013
  • DominiRican1013
I don't know how to do this.
Michele_Laino
  • Michele_Laino
more explanation: we have: \[2\left( {2{{\left( {\cos x} \right)}^2} - 1} \right) = 4{\left( {\cos x} \right)^2} - 2\] am I right?
DominiRican1013
  • DominiRican1013
these are the choices \[A. x=2\pi, x=\pi/3\] \[B. x=\pi, x=2\pi/3, x=4\pi/3\] \[C. x=2\pi, x=\pi/4, x=5\pi/4\] \[D. x=\pi/6, x=7\pi/6\]
Michele_Laino
  • Michele_Laino
yes! I see them, nevertheless I can not give you the answer directly, since I have to respect the Code of Conduct
Michele_Laino
  • Michele_Laino
If I simplify my equation above, I get: \[4{\left( {\cos x} \right)^2} + 3\cos x - 1 = 0\] now, I make this substitution: cos(x)=y, so I can write: \[4{y^2} + 3y - 1 = 0\] which is a quadratic equation. please solve that equation for y
DominiRican1013
  • DominiRican1013
B is the answer I got it

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