How do you solve cos2θ+3cosθ+2=0 ?

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using interval [0,360)

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Answer 1 · 2018-04-27T16:00:35

See below

$cos2θ+3cosθ+2=0$

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 · 2018-04-27T16:02:02

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

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

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. $theta=-120 + 360 = 240^circ.$

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