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

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline9sin⁡2θ+sin⁡θ=−2sin⁡θ+2 9 \sin ^{2} \theta+\sin \theta=-2 \sin \theta+2 \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.\newline9sin⁡2θ+sin⁡θ=−2sin⁡θ+2 9 \sin ^{2} \theta+\sin \theta=-2 \sin \theta+2 \newlineAnswer: θ= \theta=
  1. Simplify Equation: Simplify the given equation by moving all terms to one side to set the equation to zero.\newline9sin⁡2θ+sin⁡θ=−2sin⁡θ+29\sin^{2}\theta + \sin \theta = -2\sin \theta + 2\newlineAdd 2sin⁡θ2\sin \theta to both sides and subtract 22 from both sides to get:\newline9sin⁡2θ+3sin⁡θ−2=09\sin^{2}\theta + 3\sin \theta - 2 = 0
  2. Factor Quadratic Equation: Factor the quadratic equation in terms of sin⁡θ\sin \theta. We are looking for two numbers that multiply to (9)(−2)=−18(9)(-2) = -18 and add up to 33. These numbers are 66 and −3-3. So we can write the equation as: (3sin⁡θ−2)(3sin⁡θ+1)=0(3\sin \theta - 2)(3\sin \theta + 1) = 0
  3. Solve First Factor: Solve each factor for sin⁡θ\sin \theta.\newlineFirst factor: 3sin⁡θ−2=03\sin \theta - 2 = 0\newlineAdd 22 to both sides:\newline3sin⁡θ=23\sin \theta = 2\newlineDivide by 33:\newlinesin⁡θ=23\sin \theta = \frac{2}{3}
  4. Solve Second Factor: Solve the second factor for sin⁡θ\sin \theta.\newlineSecond factor: 3sin⁡θ+1=03\sin \theta + 1 = 0\newlineSubtract 11 from both sides:\newline3sin⁡θ=−13\sin \theta = -1\newlineDivide by 33:\newlinesin⁡θ=−13\sin \theta = -\frac{1}{3}
  5. Find Angle for sin⁡(23)\sin(\frac{2}{3}): Find the angles that correspond to sin⁡θ=23\sin \theta = \frac{2}{3}. Since sin⁡θ=23\sin \theta = \frac{2}{3} is not a standard angle, we will use a calculator to find the inverse sine (arcsin) of 23\frac{2}{3}. θ=arcsin⁡(23)\theta = \arcsin(\frac{2}{3}) This will give us the principal angle in the first quadrant. We also need to find the angle in the second quadrant where sine is positive. θ1=arcsin⁡(23)≈41.8∘\theta_1 = \arcsin(\frac{2}{3}) \approx 41.8^\circ θ2=180∘−θ1≈180∘−41.8∘≈138.2∘\theta_2 = 180^\circ - \theta_1 \approx 180^\circ - 41.8^\circ \approx 138.2^\circ
  6. Find Angle for sin⁡(−13)\sin(-\frac{1}{3}): Find the angles that correspond to sin⁡θ=−13\sin \theta = -\frac{1}{3}.\newlineSince sin⁡θ=−13\sin \theta = -\frac{1}{3} is also not a standard angle, we will use a calculator to find the inverse sine (arcsin) of −13-\frac{1}{3}.\newlineθ=arcsin⁡(−13)\theta = \arcsin(-\frac{1}{3})\newlineThis will give us the principal angle in the fourth quadrant. We also need to find the angle in the third quadrant where sine is negative.\newlineθ3=360°−arcsin⁡(13)≈360°−19.5°≈340.5°\theta_3 = 360° - \arcsin(\frac{1}{3}) \approx 360° - 19.5° \approx 340.5°\newlineθ4=180°+arcsin⁡(13)≈180°+19.5°≈199.5°\theta_4 = 180° + \arcsin(\frac{1}{3}) \approx 180° + 19.5° \approx 199.5°

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