How do you solve cos2θ+3cosθ+2=0 ?

#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]}

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#