Question #946b8

Transform the given equation into polynomial form using the fact that #cos(2x)=2cos^2(x)-1#

#cos(2x)-3cos(x)=0#
#2cos^2(x)-1-3cos(x)=0#
#2cos^2(x)-3cos(x)-1=0#

Write this quadratic equation in standard form by dividing by 2
#cos^2(x)-3/2cos(x)-1/2=0#

Solve the quadratic equation by completing the square. Add 1/2 to both sides
#cos^2(x)-3/2cos(x)=1/2#

Take half the coefficient of #cos(x)# and square it, then add to both sides.
#cos^2(x)-3/2cos(x)+9/16=17/16#

Factor the left-hand side.
#(cos(x)-3/4)^2=17/16#

#cos(x)-3/4=+-sqrt(17)/4#

#cos(x)=3/4+-sqrt(17)/4#

#cos^-1[cos(x)]=cos^-1[3/4+-sqrt(17)/4]+2pin# with #ninZZ#
#x=cos^-1[3/4+-sqrt(17)/4]+2pin# with #ninZZ#

One might also use the Quadratic Formula.
After the step

#2cos^2(x) - 3cosx - 1 = 0#, note that the equation is quadratic in form -- much like #2y^2-3y-1=0#.

Therefore
#cosx = (3 +-sqrt(9 - 4(2)(-1)))/2(2)#
#cosx = (3 +-sqrt(9 + 8))/4#
#cosx = (3 +-sqrt(17))/4#

Now observe that #3 + sqrt(17) > 3 + 4 > 4#.
Therefore #(3 + sqrt(17))/4 > 1#, which means that cosx never attains that value.

However #(3 - sqrt(17))/4# lies between -1 and 0. Therefore

#cosx = (3 - sqrt(17))/4# for some values of x in the second and third quadrants.

When we write #Cos^(-1)u# of any value u, we mean the angle in the first or second quadrant whose cosine is "u."

Thus #x = Cos^(-1)((3 - sqrt(17))/4)# is in Quadrant II, and corresponding to it, for every integer n, #x = Cos^(-1)((3 - sqrt(17))/4) + 2npi# is a solution to the equation.

In the third quadrant, the angle with the same reference angle as #x_1 = Cos^(-1)((3 - sqrt(17))/4)# is #x_2 = 2pi - Cos^(-1)((3 - sqrt(17))/4)#

The solutions to the equation are
#x = Cos^(-1)((3 - sqrt(17))/4) + 2npi#
and
#x = -Cos^(-1)((3 - sqrt(17))/4) + 2npi#

Since cosine is an even function, [#cos(theta) = cos(-theta)# for all # theta#], this is the sort of answer that we ought to expect.