Question

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

I think B. x=pi, x=2pi/3 and x=4pi/3 is the answer is this correct. Thank you!

Answer 1

well the quadratic equation also factors ... the equation whilst using cos(x) is a quadratic.

so

\[(\cos(x) + 1) (2\cos^2(x) - 3\cos(x) - 2) = (\cos(x) +1)(\cos(x) - 2)(2\cos(x) + 1) = 0\]

now you have all the linear factors you can find the solutions for x.

hope it helps

Answer 2

oops the factored form is

(cos(x) +1)(cos(x) -2)(2cos(x) + 1) = 0

Answer 3

Ok thank you!

Answer 4

@campbell_st so the solutions for x would be the answer?

Answer 5

@campbell_st I see the solutions on here at the bottom

http://www.wolframalpha.com/input/?i=%28cos%28x%29+%2B1%29%28cos%28x%29+-2%29%282cos%28x%29+%2B+1%29+%3D+0

But they don't look like any choices that I have?

Answer 6

well the solution to cos(x) - 2 = 0 doesn't exist so you can ignore that.

so you are looking to solve for x

cos(x) + 1 = 0

2cos(x) + 1 = 0

so you have

cos(x) = -1

cos(x) = -1/2

now find the angles that give these values.

Answer 7

@campbell_st how do I find the angles of the given values?

Answer 8

ok well its just

\[x = \cos^{-1} (-1)~~~~and~~~~ x = \cos^{-1}(-1/2)\]

Answer 9

@campbell_st

x=cos^-1(-1) = x=pi

x=cos^-1(-1/2) = x=2pi/3

So wouldn't that make the answer B. x=pi, x=2pi/3 and x=4pi

Answer 10

well there are 2 values for x = cos^-1 (-1/2) 2nd and 3rd quadrants