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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. csc^(2)theta+5csc 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.\newlinecsc⁡2θ+5csc⁡θ−6=0 \csc ^{2} \theta+5 \csc \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.\newlinecsc⁡2θ+5csc⁡θ−6=0 \csc ^{2} \theta+5 \csc \theta-6=0 \newlineAnswer: θ= \theta=
  1. Set xx as csc⁡(θ)\csc(\theta): Recognize that the given equation is a quadratic equation in terms of csc⁡(θ)\csc(\theta). Let's set x=csc⁡(θ)x = \csc(\theta) to make it look more familiar.\newlineThe equation becomes x2+5x−6=0x^2 + 5x - 6 = 0.
  2. Factor the quadratic: Factor the quadratic equation. \newline(x+6)(x−1)=0(x + 6)(x - 1) = 0
  3. Solve for x: Solve for x.\newlinex+6=0x + 6 = 0 or x−1=0x - 1 = 0\newlinex=−6x = -6 or x=1x = 1
  4. Translate to csc⁡(θ)\csc(\theta): Translate the solutions for xx back into terms of csc⁡(θ)\csc(\theta).\newlinecsc⁡(θ)=−6\csc(\theta) = -6 or csc⁡(θ)=1\csc(\theta) = 1
  5. Use sine definition: Solve for θ\theta using the definition of cosecant, which is the reciprocal of sine.\newlineFor csc⁡(θ)=1\csc(\theta) = 1, sin⁡(θ)=1csc⁡(θ)=11=1\sin(\theta) = \frac{1}{\csc(\theta)} = \frac{1}{1} = 1.\newlineFor csc⁡(θ)=−6\csc(\theta) = -6, sin⁡(θ)=1csc⁡(θ)=1−6\sin(\theta) = \frac{1}{\csc(\theta)} = \frac{1}{-6}.
  6. Find sin⁡(θ)=1\sin(\theta) = 1: Find the angles that correspond to sin⁡(θ)=1\sin(\theta) = 1.\newlinesin⁡(θ)=1\sin(\theta) = 1 at θ=90°\theta = 90°.
  7. Find sin⁡(θ)=−16\sin(\theta) = -\frac{1}{6}: Find the angles that correspond to sin⁡(θ)=1−6\sin(\theta) = \frac{1}{-6}.\newlineSince the range of the sine function is [−1,1][-1, 1], and −16-\frac{1}{6} is within this range, we look for angles where the sine is negative. This occurs in the third and fourth quadrants.
  8. Find reference angle: Use the inverse sine function to find the reference angle for sin⁡(θ)=−16\sin(\theta) = -\frac{1}{6}.θref=arcsin⁡(16)≈9.6∘\theta_{\text{ref}} = \arcsin(\frac{1}{6}) \approx 9.6^\circ
  9. Determine angles: Determine the angles in the third and fourth quadrants that have this reference angle.\newlineFor the third quadrant: θ=180°+θref≈180°+9.6°≈189.6°\theta = 180° + \theta_{\text{ref}} \approx 180° + 9.6° \approx 189.6°\newlineFor the fourth quadrant: θ=360°−θref≈360°−9.6°≈350.4°\theta = 360° - \theta_{\text{ref}} \approx 360° - 9.6° \approx 350.4°
  10. Combine solutions: Combine all the solutions. θ=90°,189.6°,350.4°\theta = 90°, 189.6°, 350.4°

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