For what fraction of each six-hour cycle does 10+2sin(πt/3) exceed 11? An inequality selects arcs and time intervals, not just crossing points. Use intervals, sinusoids and equations.
10 + 2sin(πt/3) > 11 ⟺ sinθ > 1/2, θ = πt/3.
On the unit circle sinθ is height. The line y=1/2 crosses at π/6 and 5π/6; the open upper arc lies above it. Angles are radians; k is any integer.
π/6 + 2πk < θ < 5π/6 + 2πk; 1/2 + 6k < t < 5/2 + 6k.
Open arc sinθ>1/2
x²+y²=1
Time above threshold
y=10+2sin(πt/3); -- y=11
This is 2 of every 6 hours, one third of the time. On 0≤t≤12, the answer is (0.5,2.5)∪(6.5,8.5) hours. It is a continuous hypothetical signal, not sampled sensor data. For ≥ endpoints would be included; for > they equal 11 and are excluded. OpenStax
Breaks are boundaries too
tanθ > 1 ⟺ π/4 + πk < θ < π/2 + πk.
The right endpoint is excluded because tan is undefined at π/2+πk, not because of strictness alone. Solve on one branch, then repeat every π.
Extra: intersect conditions
sinθ ≥ 1/2, cosθ > 0 ⟺ π/6 + 2πk ≤ θ < π/2 + 2πk.
Include the start because of ≥; exclude the end because cosθ=0. A system means “and”: intersect, do not unite.
Extra: a substitution with an isolated value
2(sin)^(2)θ − sinθ − 1 ≥ 0; u = sinθ ∈ [ − 1,1].
(2u + 1)(u − 1) ≥ 0 ⟺ u ∈ [ − 1, − 1/2]∪{1}.
On one turn this gives:
θ ∈ [7π/6,11π/6]∪{π/2} (0 ≤ θ < 2π).
Repeat both parts every 2π. The isolated point π/2 is a genuine non-strict solution. Check domain, one period and every endpoint before converting angle back to time.
The reverse substitution x=sin t can simplify radicals. For −1≤x≤1 choose −π/2≤t≤π/2: the correspondence is one-to-one, cos t≥0 and √(1−x²)=cos t. Then √(1−x²)>x becomes cos t>sin t, or t<π/4 within this interval; the answer is −1≤x<√2/2. Include the left endpoint and exclude the right. This requires a domain and a return substitution; it does not allow √(cos²t)=cos t everywhere.