Question

(cos x+1)(2cos^2 x-3cos x-2)?
Solve on the interval [0,2pi)


A. x=2pi, x=pi/2, x=5pi/4
B. x=pi/6,x=7pi/6
C. x=2pi,x=pi/2,x=pi/3
D. x=pi, x=2pi/3, x=4pi/3

Answer 1

Is there meant to be a "=0" at the end of that or is it just an expression?

Answer 2

lol yeah sorry about that there is an =0 at the end.

Answer 3

Okay, well you can separate into two parts if that's the case:
\[(\cos x+1)(2\cos^2 x-3\cos x-2)=0\]
\[\cos x+1=0\]
or
\[2\cos^2 x -3\cos x -2=0\]
Are you capable of factorising the one above ^?

Answer 4

no :( im sorry i dont know how to .

Answer 5

You just imagine cosx to be x. and you can factorise it like a quadratic.

Answer 6

I'm sure you must of factorised a quadratic equation before, am I right?

Answer 7

not sure :/ can you set it up for me please

Answer 8

Can you factorise this:
\[2x^2-3x-2=0\]

Answer 9

b
n