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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+sin theta=0 Answer: theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline4sin⁡2θ+sin⁡θ=0 4 \sin ^{2} \theta+\sin \theta=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.\newline4sin⁡2θ+sin⁡θ=0 4 \sin ^{2} \theta+\sin \theta=0 \newlineAnswer: θ= \theta=
  1. Factor equation: Factor the given trigonometric equation.\newlineThe equation given is 4sin⁡2(θ)+sin⁡(θ)=04\sin^2(\theta) + \sin(\theta) = 0. We can factor out sin⁡(θ)\sin(\theta) from both terms.\newlinesin⁡(θ)(4sin⁡(θ)+1)=0\sin(\theta)(4\sin(\theta) + 1) = 0
  2. Set equal and solve: Set each factor equal to zero and solve for θ\theta. We have two factors: sin⁡(θ)\sin(\theta) and (4sin⁡(θ)+1)(4\sin(\theta) + 1). Setting each equal to zero gives us two separate equations to solve: 11. sin⁡(θ)=0\sin(\theta) = 0 22. 4sin⁡(θ)+1=04\sin(\theta) + 1 = 0
  3. Solve sin⁡(θ)=0\sin(\theta) = 0: Solve the first equation sin⁡(θ)=0\sin(\theta) = 0. The sine function is equal to zero at 0∘0^\circ and 180∘180^\circ within the range of 0∘0^\circ to 360∘360^\circ. Therefore, θ=0∘\theta = 0^\circ and θ=180∘\theta = 180^\circ.
  4. Solve 4sin⁡(θ)+1=04\sin(\theta) + 1 = 0: Solve the second equation 4sin⁡(θ)+1=04\sin(\theta) + 1 = 0 for θ\theta.\newlineSubtract 11 from both sides to isolate the sine term:\newline4sin⁡(θ)=−14\sin(\theta) = -1\newlineDivide both sides by 44:\newlinesin⁡(θ)=−14\sin(\theta) = -\frac{1}{4}
  5. Find angles for sin⁡(θ)=−14\sin(\theta) = -\frac{1}{4}: Find the angles that correspond to sin⁡(θ)=−14\sin(\theta) = -\frac{1}{4}. Since the sine function is negative, we are looking for angles in the third and fourth quadrants where sine is negative. Using the inverse sine function on a calculator, we find the reference angle: θ=arcsin⁡(14)≈14.5∘\theta = \arcsin(\frac{1}{4}) \approx 14.5^\circ
  6. Determine angles in quadrants: Determine the actual angles in the third and fourth quadrants.\newlineFor the third quadrant, we add 180∘180^\circ to the reference angle:\newlineθ≈180∘+14.5∘=194.5∘\theta \approx 180^\circ + 14.5^\circ = 194.5^\circ\newlineFor the fourth quadrant, we subtract the reference angle from 360∘360^\circ:\newlineθ≈360∘−14.5∘=345.5∘\theta \approx 360^\circ - 14.5^\circ = 345.5^\circ
  7. Combine all solutions: Combine all the solutions.\newlineThe angles that satisfy the equation 4sin⁡2(θ)+sin⁡(θ)=04\sin^2(\theta) + \sin(\theta) = 0 are:\newlineθ=0∘\theta = 0^\circ, θ=180∘\theta = 180^\circ, θ=194.5∘\theta = 194.5^\circ, and θ=345.5∘\theta = 345.5^\circ.

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