Find every solution in the open interval.

01 / Standard#Your turn

\sum_{k=1}^6\sin(k\theta)=0,\quad0<\theta<2\pi

Official paper · jm02-2025 · 4(d) · PDF 6

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Take imaginary parts of the geometric sum with n=6.

Hint 2

Also check θ=π, excluded in the preceding part’s wording.

Worked solution

For θ≠π, the identity gives a product, and its denominator is nonzero.
$\sum_{k=1}^6\sin(k\theta)=\frac{\sin3\theta\sin(7\theta/2)}{\sin(\theta/2)}$
Solve each factor and retain the open interval.
$\theta=\frac{k\pi}3\ (k=1,2,3,4,5),\quad\text{or}\quad\theta=\frac{2m\pi}7\ (m=1,2,3,4,5,6)$
At θ=π every summand is zero, so it is indeed included; the two lists have no overlap in this interval.

Eleven solutions: kπ/3 (k=1,…,5) and 2mπ/7 (m=1,…,6).

Checks and common pitfalls: Do not discard θ=π from this part just because the previous part excluded it.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
For θ≠π, the identity gives a product, and its denominator is nonzero.
$\sum_{k=1}^6\sin(k\theta)=\frac{\sin3\theta\sin(7\theta/2)}{\sin(\theta/2)}$
Solve each factor and retain the open interval.
$\theta=\frac{k\pi}3\ (k=1,2,3,4,5),\quad\text{or}\quad\theta=\frac{2m\pi}7\ (m=1,2,3,4,5,6)$
At θ=π every summand is zero, so it is indeed included; the two lists have no overlap in this interval.

Think first. Reveal a hint when the class is ready.

2025 JM02