An equation is given. Let k be any integer.

Question

2/√3 cos(3θ) = 1

 

(a) Find all solutions of the equation.

 

(b) Find the solutions in the interval [0, 2π).

Answer 1

(2/√3)cos(3θ) = 1

 

          cos(3θ) = √3/2

 

3θ = π/6 + 2kπ, 11π/6 + 2kπ where k = 0, ±1, ±2, ...

 

 θ = π/18 + 2kπ/3, 11π/18 + 2kπ/3

 

If k = 0, we get θ = π/18, 11π/18

If k = 1, we get θ = 13π/18, 22π/18

If k = 2, we get θ = 25π/18, 35π/18

 

Other choices of k give values of θ that lie outside of the interval [0, 2π]

 

 

Answer 2

First thing to note is that cos(pi/6) = (√3)/2

 

2/√3 · cos(3θ) = 1

cos(3θ) = (√3)/2

3θ = pi/6

θ = pi/18   This is the solution in quadrant (Q) I.

 

In QII and QIII, the angles 5pi/6 and 7pi/6 also have a reference angle of pi/6 but the cosine is negative (= -(√3)/2)  which would give an answer of -1.  So there is no solution in QII or QIII.  In QIV, 11pi/6 has a reference angle of pi/6 and is also positive.  Hence on the interval [0,2pi] the answers are:

 

3θ = pi/6

θ = pi/18

and

3θ = 11pi/6

θ = 11pi/18

 

The full set of solutions repeats every 2pi:

 

θ = pi/18 ± (2pi)n    n = 0, 1, 2, 3, ...

θ = 11pi/18 ± (2pi)n    n = 0, 1, 2, 3, ...