Solve the equation for all degree solutions and if 0° ≤ θ < 360°. Do not use a calculator (Enter your answers as a comma-separated list. If there is no solution, write no solution.)1. 2 sin θ= root 2(a) all degree solutions (let k be any integer.)(b) 0≤ θ < 360°2. 2 cos θ − root 3=0(a) All degree solutions (let k be any integer.)(b) 0≤ θ < 360°3. root 3 cot θ- 1=0(a) all degree solutions (Let k be any integer.)(b) 0° ≤ θ < 360°
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Answer:1.- a) θ = 45 ° + 360k θ = 135° + 360kb) θ = 45 ° 0 ≤ θ < 360° θ = 135° 2.- a) θ = 30° + 360k θ = 330° + 360kb) θ = 30° 0 ≤ θ < 360° θ = 330° 3.- a) θ = 60° + 360k θ = 240° + 360kb) θ = 60° 0 ≤ θ < 360° θ = 240° Step-by-step explanation:1) 2 sin θ = √2 sin θ = √2/21.a ) θ = 45 ° + 360k θ = 135° + 360k1.b) θ = 45 ° 0 ≤ θ < 360° θ = 135° 2.2 cos θ - √3 = 0 2 cos θ = √3 cos θ = √3/22.a θ = 30° + 360k θ = 330° + 360k2.b θ = 30° 0 ≤ θ < 360° θ = 330° 3.- √3 cot θ - 1 = 0 √3 cot θ = 1 cot θ = 1 /√33.a θ = 60° + 360k θ = 240° + 360k3.b θ = 60° 0 ≤ θ < 360° θ = 240°
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10 months ago
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