Question #946b8

2 Answers
May 14, 2017

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

Explanation:

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

May 15, 2017

An alternative and a few details.

Explanation:

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.