For 0≤θ≤π/2, the displayed system has more than one solution. Find θ and its full solution.

01 / Standard#Your turn

\begin{cases}x+y+z=6\\(\sin2\theta)x+(\sin4\theta)y+(\sin8\theta)z=\sqrt3\\(\cos2\theta)x+(\cos4\theta)y+(\cos8\theta)z=-3\end{cases}

Official paper · jm02-2023 · 5(b) · PDF 7

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Nonuniqueness forces the determinant in part (a) to vanish.

Hint 2

Test consistency at each possible θ.

Worked solution

Find the candidate angles.
$\sin\theta\sin2\theta\sin3\theta=0\implies\theta=0,\pi/3,\pi/2$
At 0 and π/2 the second equation becomes 0=√3, so both are impossible.
At π/3 the cosine equation is dependent; the other two equations give y=2 and x+z=4.
$x+y+z=6,\quad x-y+z=2\implies(x,y,z)=(4-t,2,t),\quad t\in\mathbb R$

θ=π/3; (x,y,z)=(4−t,2,t), t∈ℝ.

Checks and common pitfalls: A zero determinant is necessary but consistency must still be checked.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Find the candidate angles.
$\sin\theta\sin2\theta\sin3\theta=0\implies\theta=0,\pi/3,\pi/2$
At 0 and π/2 the second equation becomes 0=√3, so both are impossible.
At π/3 the cosine equation is dependent; the other two equations give y=2 and x+z=4.
$x+y+z=6,\quad x-y+z=2\implies(x,y,z)=(4-t,2,t),\quad t\in\mathbb R$

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

2023 JM02