Question

For real numbers $x,$ let $\begin{aligned} P(x) &= 1 + \cos(x) + i\sin(x) \\ &\quad {}- \cos(2x) - i\sin(2x) \\ &\quad {}+ \cos(3x) + i\sin(3x) \end{aligned}$ where $i = \sqrt{-1}.$ For how many values of $x$ with $0 \le x \lt 2\pi$ does $P(x) = 0?$

$0$

$1$

$2$

$3$

$4$

Answer: A

Concepts:complex number trigonometric identity

Difficulty rating: 2420

Solution:

Group by Euler's formula: $P(x) = 1 + e^{ix} - e^{2ix} + e^{3ix}.$ The imaginary part is $\begin{gathered} \sin x - \sin 2x + \sin 3x \\ = (\sin x + \sin 3x) - \sin 2x \\ = \sin 2x(2\cos x - 1). \end{gathered}$

This vanishes when $\sin 2x = 0$ (so $x = 0, \tfrac{\pi}{2}, \pi, \tfrac{3\pi}{2}$ ) or $\cos x = \tfrac12$ (so $x = \tfrac{\pi}{3}, \tfrac{5\pi}{3}$ ).

Checking the real part $1 + \cos x - \cos 2x + \cos 3x$ at each of these values gives $\pm 2$ or $1,$ never $0.$ So no $x$ makes $P(x) = 0.$

Thus, the correct answer is A.

2021 AMC 12B Fall Problem 21

Answer: A

Difficulty rating: 2420

Solution:

Problem 21 in Other Years