## DominiRican1013 one year ago Solve on the interval [0,2π): 2cos2x+3cosx+1=0

1. Michele_Laino

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

2. Michele_Laino

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

3. Michele_Laino

4. Michele_Laino

what equation do you get?

5. DominiRican1013

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

6. Michele_Laino

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

7. DominiRican1013

I'll be right back.

8. DominiRican1013

I don't know how to do this.

9. 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?

10. DominiRican1013

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

11. Michele_Laino

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

12. 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

13. DominiRican1013

B is the answer I got it

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