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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. -5sin^(2)theta-3=2sin theta-6 Answer: theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline−5sin⁡2θ−3=2sin⁡θ−6 -5 \sin ^{2} \theta-3=2 \sin \theta-6 \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−5sin⁡2θ−3=2sin⁡θ−6 -5 \sin ^{2} \theta-3=2 \sin \theta-6 \newlineAnswer: θ= \theta=
  1. Rewrite equation: First, let's rewrite the given equation to have all terms on one side, which will help us to solve for θ\theta.
    −5sin⁡2(θ)−3=2sin⁡(θ)−6-5\sin^2(\theta) - 3 = 2\sin(\theta) - 6
    Add 66 to both sides and subtract 2sin⁡(θ)2\sin(\theta) from both sides to get a quadratic equation in terms of sin⁡(θ)\sin(\theta).
    −5sin⁡2(θ)−2sin⁡(θ)+3=0-5\sin^2(\theta) - 2\sin(\theta) + 3 = 0
  2. Factor quadratic equation: Next, we factor the quadratic equation to find the possible values of sin⁡(θ)\sin(\theta). Let's look for two numbers that multiply to −5×3=−15-5\times3 = -15 and add up to −2-2. These numbers are −5-5 and +3+3. So we can write the equation as: (−5sin⁡(θ)+3)(sin⁡(θ)+1)=0(-5\sin(\theta) + 3)(\sin(\theta) + 1) = 0
  3. Solve for sin⁡(θ)\sin(\theta): Now, we set each factor equal to zero and solve for sin⁡(θ)\sin(\theta). First factor: −5sin⁡(θ)+3=0-5\sin(\theta) + 3 = 0 Add 5sin⁡(θ)5\sin(\theta) to both sides: 3=5sin⁡(θ)3 = 5\sin(\theta) Divide both sides by 55: sin⁡(θ)=35\sin(\theta) = \frac{3}{5}
  4. Find angles for sin⁡(θ)\sin(\theta): Second factor: sin⁡(θ)+1=0\sin(\theta) + 1 = 0\newlineSubtract 11 from both sides:\newlinesin⁡(θ)=−1\sin(\theta) = -1
  5. Consider symmetry of sine function: Now we need to find the angles θ\theta that correspond to sin⁡(θ)=35\sin(\theta) = \frac{3}{5} and sin⁡(θ)=−1\sin(\theta) = -1 within the range 0^\circ \leq \theta < 360^\circ. For sin⁡(θ)=35\sin(\theta) = \frac{3}{5}, we use the inverse sine function to find the principal value: θ=arcsin⁡(35)\theta = \arcsin(\frac{3}{5})
  6. Calculate actual angles: The calculator gives us the principal value, but we need to consider the symmetry of the sine function to find all solutions within the given range.\newlineThe sine function is positive in the first and second quadrants. We already have the first quadrant angle from the inverse sine function. To find the second quadrant angle, we use the fact that sine is symmetric about 180°180°.\newlineθ1=180°−arcsin⁡(35)\theta_1 = 180° - \arcsin(\frac{3}{5})
  7. Final solution: For sin⁡(θ)=−1\sin(\theta) = -1, the angle θ\theta is straightforward since there is only one angle in the range 0^\circ \leq \theta < 360^\circ where the sine of the angle is −1-1.\newlineθ=270∘\theta = 270^\circ
  8. Final solution: For sin⁡(θ)=−1\sin(\theta) = -1, the angle θ\theta is straightforward since there is only one angle in the range 0^\circ \leq \theta < 360^\circ where the sine of the angle is −1-1. θ=270∘\theta = 270^\circ Now we calculate the actual angles using a calculator. θ=arcsin⁡(35)≈36.9∘\theta = \arcsin(\frac{3}{5}) \approx 36.9^\circ θ1=180∘−arcsin⁡(35)≈180∘−36.9∘≈143.1∘\theta_1 = 180^\circ - \arcsin(\frac{3}{5}) \approx 180^\circ - 36.9^\circ \approx 143.1^\circ
  9. Final solution: For sin⁡(θ)=−1\sin(\theta) = -1, the angle θ\theta is straightforward since there is only one angle in the range 0^\circ \leq \theta < 360^\circ where the sine of the angle is −1-1. θ=270∘\theta = 270^\circ Now we calculate the actual angles using a calculator. θ=arcsin⁡(35)≈36.9∘\theta = \arcsin(\frac{3}{5}) \approx 36.9^\circ θ1=180∘−arcsin⁡(35)≈180∘−36.9∘≈143.1∘\theta_1 = 180^\circ - \arcsin(\frac{3}{5}) \approx 180^\circ - 36.9^\circ \approx 143.1^\circ We have found all the angles that satisfy the given equation within the range 0^\circ \leq \theta < 360^\circ. θ≈36.9∘,143.1∘,\theta \approx 36.9^\circ, 143.1^\circ, and 270∘270^\circ

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