Solve on the stated interval.

01 / Standard#Your turn

1+\sin\theta-2\cos^2\theta=0,\quad0\le\theta<\pi

Official paper · jm01-2024 · I.13 · PDF 3

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Replace cos²θ by 1−sin²θ.

Hint 2

Sine is nonnegative on this interval.

Worked solution

Factor the quadratic in sine.
$2\sin^2\theta+\sin\theta-1=(2\sin\theta-1)(\sin\theta+1)=0$
Discard sinθ=−1 and retain both acute/supplementary angles.
$\sin\theta=1/2\implies\theta=\pi/6,5\pi/6$

Checks and common pitfalls: Do not omit the second-quadrant solution.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Factor the quadratic in sine.
$2\sin^2\theta+\sin\theta-1=(2\sin\theta-1)(\sin\theta+1)=0$
Discard sinθ=−1 and retain both acute/supplementary angles.
$\sin\theta=1/2\implies\theta=\pi/6,5\pi/6$

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

2024 JM01