Use part (i) to solve cos x+cos3x+cos5x=0 for 0≤x≤2π.

01 / Standard#Your turn

Official paper · jm02-2021 · 4(c)(ii) · PDF 6

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Set n=3 in the identity.

Hint 2

Check sin x=0 separately before dividing.

Worked solution

The product identity reduces the equation where sin x is nonzero.
$2\sin x(\cos x+\cos3x+\cos5x)=\sin6x$
At 0,π,2π the original left side is 3,−3,3, so none is a solution.
$\sin x=0\implies x=0,\pi,2\pi$
Solve sin6x=0 and exclude those three points.
$x=j\pi/6,\quad j\in\{1,2,3,4,5,7,8,9,10,11\}$

x=jπ/6 for j=1,2,3,4,5,7,8,9,10,11.

Checks and common pitfalls: Multiplying by sin x introduces candidate zeros at 0,π,2π; the original equation rejects them.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The product identity reduces the equation where sin x is nonzero.
$2\sin x(\cos x+\cos3x+\cos5x)=\sin6x$
At 0,π,2π the original left side is 3,−3,3, so none is a solution.
$\sin x=0\implies x=0,\pi,2\pi$
Solve sin6x=0 and exclude those three points.
$x=j\pi/6,\quad j\in\{1,2,3,4,5,7,8,9,10,11\}$

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

2021 JM02