How do you solve Cos(2x)cos(x)-sin(2x)sin(x)=0 over the interval 0 to 2pi?

1 Answer
Zor Shekhtman
Mar 28, 2016

Solutions are:
pi/6, pi/2, (5pi)/6, (7pi)/6, (3pi)/2, (11pi)/6

Explanation:

This is just an exercise on a simple trigonometric formula of a cosine of a sum of two angles.

As is known,
cos(a+b)=cos(a)*cos(b)-sin(a)*sin(b)

Assuming a=2x and b=x, we get:
cos(2x+x)=cos(2x)*cos(x)-sin(2x)*sin(x)

Therefore, on the left of the equation we have cos(2x+x)=cos(3x) and the given equation is equivalent to
cos(3x)=0

Cosine of an angle (in radians) is equal to 0 at angles measured pi/2+pi *n for all integer n (positive, 0 and negative).
Therefore, the common solution of our equation is
3x=pi/2+pi*n
or
x=pi/6+pi/3*n
Varying integer n from 0 forward, we get the following solutions in the interval [0,2pi]:
pi/6, pi/2, (5pi)/6, (7pi)/6, (3pi)/2, (11pi)/6

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