Problem:
How many values of θ in the interval 0<θ≤2π satisfy
1−3sinθ+5cos3θ=0?
Answer Choices:
A. 2
B. 4
C. 5
D. 6
E. 8
Solution:
Answer (D): Rewrite the equation as 3sinθ−1=5cos3θ. One can make a fairly accurate sketch of the functions on the left-hand and right-hand sides of this equation. For y=3sinx−1, move the normal sine graph with amplitude 3 down 1 unit. The graph of y=5cos3x is the graph of y=cosx stretched vertically and compressed horizontally so as to have amplitude 5 and period 32π. It can be seen that the two graphs intersect 6 times for 0<x≤2π.
OR
Let T(θ)=1−3sinθ+5cos3θ. To show that T(θ) must have at least 6 roots, note that ∣sinθ∣≤1 for all θ, so ∣1−3sinθ∣≤1+∣3sinθ∣≤1+3=4<5. Hence T(θ)>0 when cos3θ=1, and T(θ)<0 when cos3θ=−1. Thus
T(0)T(3π)T(32π)T(π)T(34π)T(35π)T(2π)>0<0>0<0>0<0, and >0
The interval (0,2π] can be partitioned into 6 subintervals each of length 3π in which T(θ) has a change of sign, and hence a zero.
To show that T has no more than 6 roots, write sinθ and cos3θ in terms of complex exponentials, so that
Let P(z)=25−23iz2+z3+23iz4+25z6. Then T(θ)=z−3P(z). Distinct θ in the interval 0<θ≤2π correspond to distinct z on the unit circle ∣z∣=1 in the complex plane. Because a polynomial can have no more roots than its degree, P has at most 6 roots, and hence T has at most 6 roots in a period. More generally, if T(θ) is a trigonometric polynomial of degree N, and T is not identically 0 , then it has at most 2N roots, counting multiplicity, because the roots of T(θ) correspond to the roots on the unit circle of an algebraic polynomial P(z) of degree 2N.