Question

Solve on the interval
[0,2π):
2cos2x+3cosx+1=0

Answer 1

here we have to make this substitution:
\[\cos \left( {2x} \right) = 2{\left( {\cos x} \right)^2} - 1\]

Answer 2

so your equation, can be rewritten as follows:
\[2\left( {2{{\left( {\cos x} \right)}^2} - 1} \right) + 3\cos x + 1 = 0\]

Answer 3

please simplify

Answer 4

what equation do you get?

Answer 5

I don't know I'm trying to work it out

Answer 6

hint:
\[4{\left( {\cos x} \right)^2} - 2 + 3\cos x + 1 = 0\]

Answer 7

I'll be right back.

Answer 8

I don't know how to do this.

Answer 9

more explanation:
we have:
\[2\left( {2{{\left( {\cos x} \right)}^2} - 1} \right) = 4{\left( {\cos x} \right)^2} - 2\]
am I right?

Answer 10

these are the choices
\[A. x=2\pi, x=\pi/3\]
\[B. x=\pi, x=2\pi/3, x=4\pi/3\]
\[C. x=2\pi, x=\pi/4, x=5\pi/4\]
\[D. x=\pi/6, x=7\pi/6\]

Answer 11

yes! I see them, nevertheless I can not give you the answer directly, since I have to respect the Code of Conduct

Answer 12

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

Answer 13

B is the answer I got it