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

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newlinecot⁡2θ−5cot⁡θ+6=0 \cot ^{2} \theta-5 \cot \theta+6=0 \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.\newlinecot⁡2θ−5cot⁡θ+6=0 \cot ^{2} \theta-5 \cot \theta+6=0 \newlineAnswer: θ= \theta=
  1. Identify Trigonometric Equation: Let's first identify the trigonometric equation we need to solve:\newlinecot⁡2(θ)−5cot⁡(θ)+6=0\cot^{2}(\theta) - 5\cot(\theta) + 6 = 0\newlineThis is a quadratic equation in terms of cot⁡(θ)\cot(\theta). To solve it, we can factor the left-hand side of the equation.
  2. Factor Quadratic Equation: We look for two numbers that multiply to 66 and add up to −5-5. These numbers are −2-2 and −3-3. So we can factor the equation as follows:\newline(cot⁡(θ)−2)(cot⁡(θ)−3)=0(\cot(\theta) - 2)(\cot(\theta) - 3) = 0
  3. Solve First Equation: Now we have two separate equations to solve:\newline11. cot⁡(θ)−2=0\cot(\theta) - 2 = 0\newline22. cot⁡(θ)−3=0\cot(\theta) - 3 = 0
  4. Find Angle with Cotangent 22: Solving the first equation, cot⁡(θ)−2=0\cot(\theta) - 2 = 0, we get:\newlinecot⁡(θ)=2\cot(\theta) = 2\newlineNow we need to find the angles θ\theta that have a cotangent of 22.
  5. Find Angle in Third Quadrant: The cotangent of an angle is the reciprocal of the tangent. So we are looking for angles where tan⁡(θ)=12\tan(\theta) = \frac{1}{2}. We can use the arctangent function to find the principal value of θ\theta: \newlineθ=arctan⁡(12)\theta = \arctan(\frac{1}{2})
  6. Solve Second Equation: Using a calculator, we find the principal value: θ≈arctan⁡(12)≈26.6\theta \approx \arctan(\frac{1}{2}) \approx 26.6 degrees However, since the cotangent function is positive in the first and third quadrants, we need to find the other angle in the third quadrant that also has the same cotangent value.
  7. Find Angle with Cotangent 33: To find the angle in the third quadrant, we add 180180 degrees to the principal value:\newlineθ≈26.6\theta \approx 26.6 degrees +180+ 180 degrees ≈206.6\approx 206.6 degrees\newlineNow we have two angles that satisfy the first equation: 26.626.6 degrees and 206.6206.6 degrees.
  8. Combine Solutions: Next, we solve the second equation, cot⁡(θ)−3=0\cot(\theta) - 3 = 0, which gives us:cot⁡(θ)=3\cot(\theta) = 3 We repeat the process to find the angles where tan⁡(θ)=13\tan(\theta) = \frac{1}{3}.
  9. Combine Solutions: Next, we solve the second equation, cot⁡(θ)−3=0\cot(\theta) - 3 = 0, which gives us:\newlinecot⁡(θ)=3\cot(\theta) = 3\newlineWe repeat the process to find the angles where tan⁡(θ)=13\tan(\theta) = \frac{1}{3}.Using the arctangent function again, we find the principal value:\newlineθ=arctan⁡(13)\theta = \arctan(\frac{1}{3})
  10. Combine Solutions: Next, we solve the second equation, cot⁡(θ)−3=0\cot(\theta) - 3 = 0, which gives us:\newlinecot⁡(θ)=3\cot(\theta) = 3\newlineWe repeat the process to find the angles where tan⁡(θ)=13\tan(\theta) = \frac{1}{3}.Using the arctangent function again, we find the principal value:\newlineθ=arctan⁡(13)\theta = \arctan(\frac{1}{3})Using a calculator, we find the principal value:\newlineθ≈arctan⁡(13)≈18.4\theta \approx \arctan(\frac{1}{3}) \approx 18.4 degrees\newlineAgain, we need to find the angle in the third quadrant with the same cotangent value.
  11. Combine Solutions: Next, we solve the second equation, cot⁡(θ)−3=0\cot(\theta) - 3 = 0, which gives us:\newlinecot⁡(θ)=3\cot(\theta) = 3\newlineWe repeat the process to find the angles where tan⁡(θ)=13\tan(\theta) = \frac{1}{3}.Using the arctangent function again, we find the principal value:\newlineθ=arctan⁡(13)\theta = \arctan(\frac{1}{3})Using a calculator, we find the principal value:\newlineθ≈arctan⁡(13)≈18.4\theta \approx \arctan(\frac{1}{3}) \approx 18.4 degrees\newlineAgain, we need to find the angle in the third quadrant with the same cotangent value.To find the angle in the third quadrant, we add 180180 degrees to the principal value:\newlineθ≈18.4\theta \approx 18.4 degrees +180+ 180 degrees ≈198.4\approx 198.4 degrees\newlineNow we have two more angles that satisfy the second equation: 18.418.4 degrees and cot⁡(θ)=3\cot(\theta) = 300 degrees.
  12. Combine Solutions: Next, we solve the second equation, cot⁡(θ)−3=0\cot(\theta) - 3 = 0, which gives us:\newlinecot⁡(θ)=3\cot(\theta) = 3\newlineWe repeat the process to find the angles where tan⁡(θ)=13\tan(\theta) = \frac{1}{3}.Using the arctangent function again, we find the principal value:\newlineθ=arctan⁡(13)\theta = \arctan(\frac{1}{3})Using a calculator, we find the principal value:\newlineθ≈arctan⁡(13)≈18.4\theta \approx \arctan(\frac{1}{3}) \approx 18.4 degrees\newlineAgain, we need to find the angle in the third quadrant with the same cotangent value.To find the angle in the third quadrant, we add 180180 degrees to the principal value:\newlineθ≈18.4\theta \approx 18.4 degrees +180+ 180 degrees ≈198.4\approx 198.4 degrees\newlineNow we have two more angles that satisfy the second equation: 18.418.4 degrees and cot⁡(θ)=3\cot(\theta) = 300 degrees.Combining the solutions from both equations, we have four angles that satisfy the original equation:\newlinecot⁡(θ)=3\cot(\theta) = 311 degrees, cot⁡(θ)=3\cot(\theta) = 322 degrees, 18.418.4 degrees, and cot⁡(θ)=3\cot(\theta) = 300 degrees.\newlineThese are the angles between cot⁡(θ)=3\cot(\theta) = 355 and cot⁡(θ)=3\cot(\theta) = 366 degrees that satisfy the given trigonometric equation to the nearest tenth of a degree.

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