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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. -4cos^(2)theta+13 cos theta-5=9cos theta-8 Answer: theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline−4cos⁡2θ+13cos⁡θ−5=9cos⁡θ−8 -4 \cos ^{2} \theta+13 \cos \theta-5=9 \cos \theta-8 \newlineAnswer: θ= \theta=

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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−4cos⁡2θ+13cos⁡θ−5=9cos⁡θ−8 -4 \cos ^{2} \theta+13 \cos \theta-5=9 \cos \theta-8 \newlineAnswer: θ= \theta=
  1. Rewrite Equation: First, let's rewrite the given equation to collect like terms on one side:\newline−4cos⁡2(θ)+13cos⁡(θ)−5=9cos⁡(θ)−8-4\cos^2(\theta) + 13\cos(\theta) - 5 = 9\cos(\theta) - 8
  2. Combine Like Terms: Now, subtract 9cos⁡(θ)9\cos(\theta) from both sides and add 88 to both sides to get all terms involving cos⁡(θ)\cos(\theta) on one side and the constant on the other side:\newline−4cos⁡2(θ)+13cos⁡(θ)−5−9cos⁡(θ)+8=0-4\cos^2(\theta) + 13\cos(\theta) - 5 - 9\cos(\theta) + 8 = 0
  3. Quadratic Equation: Simplify the equation by combining like terms: \newline−4cos⁡2(θ)+4cos⁡(θ)+3=0-4\cos^2(\theta) + 4\cos(\theta) + 3 = 0
  4. Factorize: This is a quadratic equation in terms of cos⁡(θ)\cos(\theta). Let's set u=cos⁡(θ)u = \cos(\theta) to make it easier to solve:\newline−4u2+4u+3=0-4u^2 + 4u + 3 = 0
  5. Solve for uu: Now, we can factor the quadratic equation:\newline−4u2+4u+3=0-4u^2 + 4u + 3 = 0\newline(2u+1)(−2u+3)=0(2u + 1)(-2u + 3) = 0
  6. Find Possible Values: Set each factor equal to zero and solve for uu:2u+1=02u + 1 = 0 or −2u+3=0-2u + 3 = 0
  7. Discard Invalid Value: Solve the first equation for uu:2u+1=02u + 1 = 02u=−12u = -1u=−12u = -\frac{1}{2}
  8. Find Corresponding Angles: Solve the second equation for uu:\(\newline\)−2-2uu + 33 = 00\(\newline\)−2-2uu = −3-3\(\newline\)uu = 32\frac{3}{2}
  9. Final Answer: Since u=cos⁡(θ)u = \cos(\theta), we have two possible values for cos⁡(θ)\cos(\theta): −12-\frac{1}{2} and 32\frac{3}{2}. However, the cosine of an angle cannot be greater than 11, so we discard the value 32\frac{3}{2}.
  10. Final Answer: Since u=cos⁡(θ)u = \cos(\theta), we have two possible values for cos⁡(θ)\cos(\theta): −12-\frac{1}{2} and 32\frac{3}{2}. However, the cosine of an angle cannot be greater than 11, so we discard the value 32\frac{3}{2}.Now we find the angles θ\theta that correspond to cos⁡(θ)=−12\cos(\theta) = -\frac{1}{2}. The cosine of −12-\frac{1}{2} occurs at 120120 degrees and cos⁡(θ)\cos(\theta)00 degrees in the range of cos⁡(θ)\cos(\theta)11 to cos⁡(θ)\cos(\theta)22 degrees.
  11. Final Answer: Since u=cos⁡(θ)u = \cos(\theta), we have two possible values for cos⁡(θ)\cos(\theta): −12-\frac{1}{2} and 32\frac{3}{2}. However, the cosine of an angle cannot be greater than 11, so we discard the value 32\frac{3}{2}.Now we find the angles θ\theta that correspond to cos⁡(θ)=−12\cos(\theta) = -\frac{1}{2}. The cosine of −12-\frac{1}{2} occurs at 120120 degrees and cos⁡(θ)\cos(\theta)00 degrees in the range of cos⁡(θ)\cos(\theta)11 to cos⁡(θ)\cos(\theta)22 degrees.We can now write the final answer with the angles that satisfy the original equation:\newlinecos⁡(θ)\cos(\theta)33 degrees, cos⁡(θ)\cos(\theta)00 degrees

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