How do you solve 2+ cos(2x)=3 cos (x) over the interval 0 to 2pi?

1 Answer
Feb 1, 2016

x=0, pi/3

Explanation:

We want a function in terms of either cos(x) or cos( 2x), but not both. We can use the double angle formula to convert cos(2x) into an expression with cos(x).

Double Angle Formula
cos(2x) = 2cos^2(x) -1

Applying the double angle formula to our function gives;

2+2cos^2(x) -1 = 3cos(x)

Or, after a little rearranging;

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

We can use the quadratic formula to factor this expression.

Quadratic Formula
(-b+-sqrt(b^2-4ac))/(2a) where ax^2 + bx +c = 0

Plugging in values for our function;

cos(x)=(3+-sqrt((-3)^2-4(2)(1)))/(2(2))

cos(x)=(3+-sqrt(9-8))/4

cos(x)=(3+-sqrt(1))/4

cos(x)=1, 1/2

A quick glance at a unit circle will show that;

cos(0)=1
cos(pi/3)=1/2

So;

x=0, pi/3