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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. 9cos^(2)theta+3cos theta=8cos theta+4 Answer: theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline9cos⁡2θ+3cos⁡θ=8cos⁡θ+4 9 \cos ^{2} \theta+3 \cos \theta=8 \cos \theta+4 \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.\newline9cos⁡2θ+3cos⁡θ=8cos⁡θ+4 9 \cos ^{2} \theta+3 \cos \theta=8 \cos \theta+4 \newlineAnswer: θ= \theta=
  1. Rewrite Equation: Rewrite the given equation to collect all terms on one side and set the equation to zero.\newline9cos⁡2(θ)+3cos⁡(θ)−8cos⁡(θ)−4=09\cos^2(\theta) + 3\cos(\theta) - 8\cos(\theta) - 4 = 0\newlineSimplify the equation by combining like terms.\newline9cos⁡2(θ)−5cos⁡(θ)−4=09\cos^2(\theta) - 5\cos(\theta) - 4 = 0
  2. Simplify Equation: Factor the quadratic equation in terms of cos⁡(θ)\cos(\theta).$9cos⁡2(θ)−5cos⁡(θ)−4\$9\cos^2(\theta) - 5\cos(\theta) - 4 = 3cos⁡(θ)+13\cos(\theta) + 13cos⁡(θ)−43\cos(\theta) - 4\)Set each factor equal to zero to find the values of cos⁡(θ)\cos(\theta).3cos⁡(θ)+1=03\cos(\theta) + 1 = 0 and 3cos⁡(θ)−4=03\cos(\theta) - 4 = 0
  3. Factor Quadratic Equation: Solve each equation for cos⁡(θ)\cos(\theta). For 3cos⁡(θ)+1=03\cos(\theta) + 1 = 0: cos⁡(θ)=−13\cos(\theta) = -\frac{1}{3} For 3cos⁡(θ)−4=03\cos(\theta) - 4 = 0: cos⁡(θ)=43\cos(\theta) = \frac{4}{3} However, since the cosine of an angle cannot be greater than 11, cos⁡(θ)=43\cos(\theta) = \frac{4}{3} is not a valid solution.
  4. Solve for cos⁡(θ)\cos(\theta): Find the angles θ\theta that correspond to cos⁡(θ)=−13\cos(\theta) = -\frac{1}{3} in the range 0^\circ \leq \theta < 360^\circ. Use the inverse cosine function and consider the symmetry of the cosine function, which is positive in the first and fourth quadrants and negative in the second and third quadrants. θ=cos⁡−1(−13)\theta = \cos^{-1}(-\frac{1}{3}) Calculate the principal value. θ≈cos⁡−1(−13)≈109.5∘\theta \approx \cos^{-1}(-\frac{1}{3}) \approx 109.5^\circ
  5. Find θ\theta for cos⁡(θ)=−13\cos(\theta) = -\frac{1}{3}: Find the second angle in the range 0^\circ \leq \theta < 360^\circ that also has a cosine of −13-\frac{1}{3}.\newlineSince cosine is negative in the second and third quadrants, the second angle is the supplement of the principal value.\newlineθ=180∘+(180∘−109.5∘)\theta = 180^\circ + (180^\circ - 109.5^\circ)\newlineθ≈180∘+70.5∘\theta \approx 180^\circ + 70.5^\circ\newlineθ≈250.5∘\theta \approx 250.5^\circ

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