Solving trigonometric equations
Problem 13.80 · medium
Solve \( \displaystyle \sin{\left(2 \theta \right)} = 1 \) for \( \displaystyle 0 \le \theta < 2\pi \).
- Let u = 2θ. Since 0 ≤ θ < 2π, u runs over [0, 4π): two full turns, so twice as many solutions.
- \[ \left. \sin{\left(2 \theta \right)} - 1 \right|_{\substack{ \theta=\frac{\pi}{4} }} = 0 \]θ = pi/4 works.✓ Proved
- \[ \left. \sin{\left(2 \theta \right)} - 1 \right|_{\substack{ \theta=\frac{5 \pi}{4} }} = 0 \]θ = 5*pi/4 works.✓ Proved
Answer \( \theta = \frac{\pi}{4},\ \frac{5 \pi}{4} \)
Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | Newton's method from 1440 starting points finds exactly 2 roots in [0, 2π), the same ones |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution is incomplete. It only lists two solutions, but there are four solutions in the interval [0, 2π) (π/4, 5π/4, 3π/4, 7π/4). The setup correctly identifies that there should be twice as many solutions, but the final answer fails to include them.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The solution is incomplete. It only lists two solutions, but there are four solutions in the interval [0, 2π) (π/4, 5π/4, 3π/4, 7π/4). The setup correctly identifies that there should be twice as many solutions, but the final answer fails to include them.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution is incomplete and misleading. It only verifies two specific values without deriving them or checking for other solutions in the range [0, 4pi) for u. Since sin(u)=1 has solutions at u=pi/2 and u=5pi/2 within [0, 4pi), the corresponding theta values are pi/4 and 5pi/4. While the final answer happens to be correct, the method shown (just checking two points) does not constitute a solution and fails to demonstrate how to find all solutions or why there are no others. A student would learn that checking random points is sufficient to solve an equation, which is false.gpt-oss:20b: pass 2026-10-14qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution is incomplete because it fails to find all solutions in the interval [0, 2π). Specifically, it misses θ = 5π/4 is a solution but θ = π/4 + π/2 = 3π/4 and θ = 5π/4 + π/2 = 7π/4 are also solutions. The equation sin(2θ)=1 implies 2θ = π/2 + 2kπ, so θ = π/4 + kπ. For k=0,1,2,3, we get π/4, 5π/4, 9π/4 (out), 13π/4 (out). Wait, let me re-evaluate. sin(2θ)=1 => 2θ = π/2 + 2nπ => θ = π/4 + nπ. In [0, 2π), n=0 gives π/4, n=1 gives 5π/4. Are there others? sin(2θ)=1. 2θ must be π/2, 5π/2, 9π/2... 2θ ∈ [0, 4π). 2θ = π/2 => θ=π/4. 2θ = 5π/2 => θ=5π/4. 2θ = 9π/2 => θ=9π/4 > 2π. So actually the stated answer IS correct. Let me re-read carefully. sin(2θ)=1. General solution 2θ = π/2 + 2kπ. θ = π/4 + kπ. Range 0 ≤ θ < 2π. k=0 -> π/4. k=1 -> 5π/4. k=2 -> 9π/4 (too big). k=-1 -> -3π/4 (too small). So the answer is correct. Why did I think 3π/4? sin(2*3π/4) = sin(3π/2) = -1. So 3π/4 is not a solution. The solution provided is actually correct. However, the reasoning in line 1 says 'u runs over [0, 4π): two full turns, so twice as many solutions.' This is vague. It doesn't explicitly derive the solutions. It just checks two values. It doesn't prove these are the ONLY solutions. A student might think checking two values is sufficient. The method is flawed/incomplete as a proof. It asserts the answer without deriving it or proving uniqueness. It just verifies two points. It misses the step of finding ALL solutions. It is misleading because it implies verification is sufficient for solving. It fails to demonstrate how to find the solutions systematically.gpt-oss:20b: pass 2026-10-14
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/trig_equations, checked 2026-10-17 with SymPy 1.14.0.