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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. tan^(2)theta+5tan theta+6=0 Answer: theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newlinetan⁡2θ+5tan⁡θ+6=0 \tan ^{2} \theta+5 \tan \theta+6=0 \newlineAnswer: θ= \theta=

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Csc, sec, and cot of special anglesright chevron icon

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Q. Find all angles, 0∘≤θ<360∘ 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newlinetan⁡2θ+5tan⁡θ+6=0 \tan ^{2} \theta+5 \tan \theta+6=0 \newlineAnswer: θ= \theta=
  1. Factor Quadratic Equation: Recognize that the given equation is a quadratic equation in terms of tan⁡(θ)\tan(\theta). We can solve for tan⁡(θ)\tan(\theta) by factoring the quadratic equation.tan⁡2(θ)+5tan⁡(θ)+6=0\tan^2(\theta) + 5\tan(\theta) + 6 = 0This can be factored into:(tan⁡(θ)+2)(tan⁡(θ)+3)=0(\tan(\theta) + 2)(\tan(\theta) + 3) = 0
  2. Solve for tan⁡(θ)\tan(\theta): Set each factor equal to zero and solve for tan⁡(θ)\tan(\theta).\newlinetan⁡(θ)+2=0\tan(\theta) + 2 = 0 or tan⁡(θ)+3=0\tan(\theta) + 3 = 0\newlineThis gives us two possible solutions for tan⁡(θ)\tan(\theta):\newlinetan⁡(θ)=−2\tan(\theta) = -2 or tan⁡(θ)=−3\tan(\theta) = -3
  3. Find tan⁡(θ)=−2\tan(\theta) = -2 angles: Find the angles that correspond to tan⁡(θ)=−2\tan(\theta) = -2 in the range 0° \leq \theta < 360°. The arctangent of −2-2 does not correspond to a special angle, so we will use a calculator to find the reference angle. arctan⁡(−2)≈−63.4°\arctan(-2) \approx -63.4° Since the tangent function is negative in the second and fourth quadrants, we find the angles in those quadrants: 180°−63.4°≈116.6°180° - 63.4° \approx 116.6° (second quadrant) 360°−63.4°≈296.6°360° - 63.4° \approx 296.6° (fourth quadrant)
  4. Find tan⁡(θ)=−3\tan(\theta) = -3 angles: Find the angles that correspond to tan⁡(θ)=−3\tan(\theta) = -3 in the range 0° \leq \theta < 360°. Again, the arctangent of −3-3 does not correspond to a special angle, so we will use a calculator to find the reference angle. arctan⁡(−3)≈−71.6°\arctan(-3) \approx -71.6° Since the tangent function is negative in the second and fourth quadrants, we find the angles in those quadrants: 180°−71.6°≈108.4°180° - 71.6° \approx 108.4° (second quadrant) 360°−71.6°≈288.4°360° - 71.6° \approx 288.4° (fourth quadrant)

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