Question

How do you solve cos2θ+3cosθ+2=0 ?

using interval [0,360)

Answer 1

See below

Explanation:

`cos2θ+3cosθ+2=0`

Apply cosine double angle identity:

`(2cos^2theta-1)+3costheta+2=0`

`2cos^2theta+3costheta+1=0`

`2cos^2theta+2costheta+costheta+1=0`

`2costheta(costheta+1)+1(costheta+1)=0`

`(2costheta+1)(costheta+1)=0`

`costheta=-1/2`
`theta= 120^@, 240^@`

`costheta=-1`
`theta= 180^@`

graph{cos(2x)+3cosx+2 [-10, 10, -5, 5]}

Answer 2

Using the double angle formula we massage this into forms `cos theta = cos a` and get

`theta = \pm 120^circ + 360^circ k or theta = 180^circ + 360^circ k`

Explanation:

The double angle formula for cosine is

`cos (2 theta ) = 2 cos^2 theta - 1`

`cos(2 theta) + 3 cos theta + 2 = 0`

`2 cos^2 theta + 3 cos theta + 1 = 0`

`(2 cos theta + 1)(cos theta + 1) = 0`

`cos theta = -1/2` or `cos theta = -1`

We got this far, don't mess up now. Remember `cos x= cos a` has solutions `x = \pm a + 360^circ k` for integer `k`.

`cos theta = cos 120^circ or cos theta = cos (180^circ )`

`theta = \pm 120^circ + 360^circ k or theta = \pm 180^circ + 360^circ k`

The `pm` doesn't really help on the `180^circ` so we land on

`theta = \pm 120^circ + 360^circ k or theta = 180^circ + 360^circ k`

Check:

Let's check one and leave the general check to you. `theta=-120 + 360 = 240^circ.`

`cos (2(240)) + 3 cos(240) + 2 = cos(120) + 3 cos(240) + 2 = -1/2 + 3(-1/2) + 2 = 0 quad sqrt`