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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. sin^(2)theta-8sin theta+12=0 Answer: theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newlinesin⁡2θ−8sin⁡θ+12=0 \sin ^{2} \theta-8 \sin \theta+12=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.\newlinesin⁡2θ−8sin⁡θ+12=0 \sin ^{2} \theta-8 \sin \theta+12=0 \newlineAnswer: θ= \theta=
  1. Rewrite as Quadratic: Recognize that the given equation is a quadratic in terms of sin⁡(θ)\sin(\theta). We can rewrite the equation as:\newline(sin⁡(θ))2−8sin⁡(θ)+12=0(\sin(\theta))^2 - 8\sin(\theta) + 12 = 0\newlineLet's set sin⁡(θ)=x\sin(\theta) = x for easier manipulation, so we have:\newlinex2−8x+12=0x^2 - 8x + 12 = 0
  2. Factor the Equation: Factor the quadratic equation.\newline(x−6)(x−2)=0(x - 6)(x - 2) = 0
  3. Solve for x: Solve for x from the factored form.\newlinex−6=0x - 6 = 0 or x−2=0x - 2 = 0\newlinex=6x = 6 or x=2x = 2\newlineHowever, since the sine function has a range of [−1,1][-1, 1], x=6x = 6 is not a valid solution for sin⁡(θ)\sin(\theta). Therefore, we only consider x=2x = 2, but again, since the sine function cannot be greater than 11, there are no solutions for this equation.

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