3.08F - Solving Sinusoidal Inequalities | Lecture Note

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High School
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High School Physics
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2024
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Chapter 3: Trigonometry Lessons 3.08F Solving Sinusoidal Inequalities Logic Period Solving Inequalities The process is the same Remember to visualize This one, you need a more accurate visual Example: −5 ≥ 2 sin 3𝑥 − 𝑃= 2𝜋 3 𝐴=2 𝜋 4 −4 𝜋 −5 ≥ 2 sin 3 𝑥 − 12 ℎ −4 ℎ+𝑃 𝑚𝑎𝑥 = −2 𝑦 = −4 −𝟓 𝑚𝑖𝑛 = −6 𝜋 12 9𝜋 12 Example: −5 ≥ 2 sin 3𝑥 − 𝜋 −5 = 2 sin 3 𝑥 − 12 −4 1 𝜋 − = sin 3 𝑥 − 2 12 𝜋 θ=3 𝑥− 12 1 − = sin(θ) 2 7π θ1 = 6 11π θ2 = 6 𝜋 4 −4 𝜋 7𝜋 3 𝑥1 − = 12 6 𝜋 7𝜋 𝑥1 − = 12 18 17𝜋 𝑥1 = 36 2𝜋 𝑃= 3 𝜋 11𝜋 3 𝑥2 − = 12 6 𝜋 11𝜋 and 𝑥2 − = 12 18 25𝜋 and 𝑥2 = 36 𝑉𝑒𝑟𝑖𝑓𝑦 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 Example: −5 ≥ 2 sin 3𝑥 − 𝑃= 2𝜋 3 𝐴=2 𝜋 4 −4 𝜋 −5 ≥ 2 sin 3 𝑥 − 12 ℎ −4 ℎ+𝑃 𝑚𝑎𝑥 = −2 𝑦 = −4 −𝟓 𝑚𝑖𝑛 = −6 𝜋 12 9𝜋 12 17𝜋 𝑥1 = 36 25𝜋 and 𝑥2 = 36 Example: −5 ≥ 2 sin 3𝑥 − 17𝜋 𝑥1 = 36 𝜋 4 −4 2𝜋 𝑃= 3 25𝜋 and 𝑥2 = 36 17𝜋 2𝜋 25𝜋 2𝜋 𝑓 ≤ −5 𝑜𝑣𝑒𝑟 + 𝑛, + 𝑛 36 3 36 3 𝑤ℎ𝑒𝑟𝑒 𝑛 ∈ ℤ Example: 2 cos 5𝑥 − 𝑃= 2𝜋 5 𝐴=2 5𝜋 8 − 4 > −3 𝜋 2 cos 5 𝑥 − 8 ℎ − 4 > −3 ℎ+𝑃 𝑚𝑎𝑥 = −2 −3 𝑦 = −4 𝑚𝑖𝑛 = −6 𝜋 8 21𝜋 40 Example: 2 cos 5 𝑥 − 𝜋 2 cos 5 𝑥 − 8 − 4 = −3 𝜋 cos 5 𝑥 − 8 1 = 2 𝜋 θ=5 𝑥− 8 𝜋 8 − 4 > −3 5 𝑥1 − 𝜋 𝜋 = 8 3 𝜋 𝜋 𝑥1 − = 8 15 23𝜋 𝑥1 = 120 2𝜋 𝑃= 5 𝜋 5𝜋 5 𝑥2 − = 8 3 𝜋 𝜋 and 𝑥2 − = 8 3 11𝜋 and 𝑥2 = 24 1 cos 𝜃 = 2 π θ1 = 3 5π θ2 = 3 𝑉𝑒𝑟𝑖𝑓𝑦 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 Example: 2 cos 5𝑥 − 𝑃= 2𝜋 5 Function is 2 𝐴= GREATER than -3, we need to ADD the period to 𝑥1 5𝜋 8 − 4 > −3 𝜋 2 cos 5 𝑥 − 8 − 4 > −3 ℎ ℎ+𝑃 𝑚𝑎𝑥 = −2 23𝜋 2𝜋 71𝜋 + = 120 5 120 𝑦 = −4 −3 𝑚𝑖𝑛 = −6 𝜋 8 21𝜋 40 23𝜋 𝑥1 = 120 11𝜋 and 𝑥2 = 24 Example: 2 cos 5 𝑥 − 11𝜋 𝑥2 = 24 𝜋 8 − 4 > −3 2𝜋 𝑃= 5 71𝜋 and 𝑥3 = 120 11𝜋 2𝜋𝑛 71𝜋 2𝜋𝑛 𝑓 > −3 𝑜𝑣𝑒𝑟 + , + 24 5 120 5 𝑤ℎ𝑒𝑟𝑒 𝑛 ∈ ℤ HW Visions 2 P119-120: 7abd, 8abd