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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. -4sin^(2)theta-6sin theta+4=-9sin theta+3 Answer: theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline−4sin⁡2θ−6sin⁡θ+4=−9sin⁡θ+3 -4 \sin ^{2} \theta-6 \sin \theta+4=-9 \sin \theta+3 \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.\newline−4sin⁡2θ−6sin⁡θ+4=−9sin⁡θ+3 -4 \sin ^{2} \theta-6 \sin \theta+4=-9 \sin \theta+3 \newlineAnswer: θ= \theta=
  1. Simplify Equation: First, we need to simplify the given equation by moving all terms to one side to set the equation to zero.\newline−4sin⁡2(θ)−6sin⁡(θ)+4=−9sin⁡(θ)+3-4\sin^2(\theta) - 6\sin(\theta) + 4 = -9\sin(\theta) + 3\newlineAdd 9sin⁡(θ)9\sin(\theta) to both sides and subtract 33 from both sides to get:\newline−4sin⁡2(θ)+3sin⁡(θ)+1=0-4\sin^2(\theta) + 3\sin(\theta) + 1 = 0
  2. Factor Quadratic Equation: Next, we factor the quadratic equation in terms of sin⁡(θ)\sin(\theta). We look for two numbers that multiply to −4-4 and add up to 33. These numbers are 44 and −1-1. So we can write the equation as: (−4sin⁡(θ)−1)(sin⁡(θ)−1)=0(-4\sin(\theta) - 1)(\sin(\theta) - 1) = 0
  3. Solve for sin⁡(θ)\sin(\theta): Now, we solve for sin⁡(θ)\sin(\theta) by setting each factor equal to zero.\newlineFirst factor: −4sin⁡(θ)−1=0-4\sin(\theta) - 1 = 0\newlineAdd 11 to both sides and then divide by −4-4:\newlinesin⁡(θ)=−14\sin(\theta) = -\frac{1}{4}
  4. Find Angles: Second factor: sin⁡(θ)−1=0\sin(\theta) - 1 = 0\newlineAdd 11 to both sides:\newlinesin⁡(θ)=1\sin(\theta) = 1
  5. Reference Angle: We now find the angles that correspond to sin⁡(θ)=−14\sin(\theta) = -\frac{1}{4} and sin⁡(θ)=1\sin(\theta) = 1 within the range 00 degrees to 360360 degrees.\newlineFor sin⁡(θ)=1\sin(\theta) = 1, the angle is 9090 degrees.
  6. Calculate Angles: For sin⁡(θ)=−14\sin(\theta) = -\frac{1}{4}, we need to find the reference angle whose sine is 14\frac{1}{4} and then find the corresponding angles in the third and fourth quadrants where sine is negative.\newlineThe reference angle, which we'll call α\alpha, can be found using the inverse sine function:\newlineα=arcsin⁡(14)≈14.5\alpha = \arcsin(\frac{1}{4}) \approx 14.5 degrees (to the nearest tenth)
  7. Final Angles: The angles in the third and fourth quadrants corresponding to the reference angle α\alpha are:\newline180+α180 + \alpha and 360−α360 - \alpha.\newlineSo we have:\newlineθ1=180+14.5≈194.5\theta_1 = 180 + 14.5 \approx 194.5 degrees\newlineθ2=360−14.5≈345.5\theta_2 = 360 - 14.5 \approx 345.5 degrees
  8. Final Angles: The angles in the third and fourth quadrants corresponding to the reference angle α\alpha are: 180+α180 + \alpha and 360−α360 - \alpha. So we have: θ1=180+14.5≈194.5\theta_1 = 180 + 14.5 \approx 194.5 degrees θ2=360−14.5≈345.5\theta_2 = 360 - 14.5 \approx 345.5 degrees We now have all the angles that satisfy the original equation: θ=90\theta = 90 degrees, 194.5194.5 degrees, and 345.5345.5 degrees.

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