Find all angles, $0^{\circ} \leq \theta<360^{\circ}$ , that satisfy the equation below, to the nearest tenth of a degree. $\newline$ $\cot ^{2} \theta-5 \cot \theta+6=0$ $\newline$ Answer: $\theta=$

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Q. Find all angles,

0^{\circ} \leq \theta<360^{\circ}

, that satisfy the equation below, to the nearest tenth of a degree.

\newline

\cot ^{2} \theta-5 \cot \theta+6=0

\newline

Answer:

\theta=

Identify Trigonometric Equation: Let's first identify the trigonometric equation we need to solve: $\newline$ $\cot^{2}(\theta) - 5\cot(\theta) + 6 = 0$ $\newline$ This is a quadratic equation in terms of $\cot(\theta)$ . To solve it, we can factor the left-hand side of the equation.
Factor Quadratic Equation: We look for two numbers that multiply to $6$ and add up to $-5$ . These numbers are $-2$ and $-3$ . So we can factor the equation as follows: $\newline$ $(\cot(\theta) - 2)(\cot(\theta) - 3) = 0$
Solve First Equation: Now we have two separate equations to solve: $\newline$ $1$ . $\cot(\theta) - 2 = 0$ $\newline$ $2$ . $\cot(\theta) - 3 = 0$
Find Angle with Cotangent $2$ : Solving the first equation, $\cot(\theta) - 2 = 0$ , we get: $\newline$ $\cot(\theta) = 2$ $\newline$ Now we need to find the angles $\theta$ that have a cotangent of $2$ .
Find Angle in Third Quadrant: The cotangent of an angle is the reciprocal of the tangent. So we are looking for angles where $\tan(\theta) = \frac{1}{2}$ . We can use the arctangent function to find the principal value of $\theta$ : $\newline$ $\theta = \arctan(\frac{1}{2})$
Solve Second Equation: Using a calculator, we find the principal value: $\theta \approx \arctan(\frac{1}{2}) \approx 26.6$ 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.
Find Angle with Cotangent $3$ : To find the angle in the third quadrant, we add $180$ degrees to the principal value: $\newline$ $\theta \approx 26.6$ degrees $+ 180$ degrees $\approx 206.6$ degrees $\newline$ Now we have two angles that satisfy the first equation: $26.6$ degrees and $206.6$ degrees.
Combine Solutions: Next, we solve the second equation, $\cot(\theta) - 3 = 0$ , which gives us: $\cot(\theta) = 3$ We repeat the process to find the angles where $\tan(\theta) = \frac{1}{3}$ .
Combine Solutions: Next, we solve the second equation, $\cot(\theta) - 3 = 0$ , which gives us: $\newline$ $\cot(\theta) = 3$ $\newline$ We repeat the process to find the angles where $\tan(\theta) = \frac{1}{3}$ .Using the arctangent function again, we find the principal value: $\newline$ $\theta = \arctan(\frac{1}{3})$
Combine Solutions: Next, we solve the second equation, $\cot(\theta) - 3 = 0$ , which gives us: $\newline$ $\cot(\theta) = 3$ $\newline$ We repeat the process to find the angles where $\tan(\theta) = \frac{1}{3}$ .Using the arctangent function again, we find the principal value: $\newline$ $\theta = \arctan(\frac{1}{3})$ Using a calculator, we find the principal value: $\newline$ $\theta \approx \arctan(\frac{1}{3}) \approx 18.4$ degrees $\newline$ Again, we need to find the angle in the third quadrant with the same cotangent value.
Combine Solutions: Next, we solve the second equation, $\cot(\theta) - 3 = 0$ , which gives us: $\newline$ $\cot(\theta) = 3$ $\newline$ We repeat the process to find the angles where $\tan(\theta) = \frac{1}{3}$ .Using the arctangent function again, we find the principal value: $\newline$ $\theta = \arctan(\frac{1}{3})$ Using a calculator, we find the principal value: $\newline$ $\theta \approx \arctan(\frac{1}{3}) \approx 18.4$ degrees $\newline$ Again, 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 $180$ degrees to the principal value: $\newline$ $\theta \approx 18.4$ degrees $+ 180$ degrees $\approx 198.4$ degrees $\newline$ Now we have two more angles that satisfy the second equation: $18.4$ degrees and $\cot(\theta) = 3$ $0$ degrees.
Combine Solutions: Next, we solve the second equation, $\cot(\theta) - 3 = 0$ , which gives us: $\newline$ $\cot(\theta) = 3$ $\newline$ We repeat the process to find the angles where $\tan(\theta) = \frac{1}{3}$ .Using the arctangent function again, we find the principal value: $\newline$ $\theta = \arctan(\frac{1}{3})$ Using a calculator, we find the principal value: $\newline$ $\theta \approx \arctan(\frac{1}{3}) \approx 18.4$ degrees $\newline$ Again, 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 $180$ degrees to the principal value: $\newline$ $\theta \approx 18.4$ degrees $+ 180$ degrees $\approx 198.4$ degrees $\newline$ Now we have two more angles that satisfy the second equation: $18.4$ degrees and $\cot(\theta) = 3$ $0$ degrees.Combining the solutions from both equations, we have four angles that satisfy the original equation: $\newline$ $\cot(\theta) = 3$ $1$ degrees, $\cot(\theta) = 3$ $2$ degrees, $18.4$ degrees, and $\cot(\theta) = 3$ $0$ degrees. $\newline$ These are the angles between $\cot(\theta) = 3$ $5$ and $\cot(\theta) = 3$ $6$ degrees that satisfy the given trigonometric equation to the nearest tenth of a degree.

Find all angles, 0∘≤θ<360∘ 0^{\circ} \leq \theta<360^{\circ} 0∘≤θ<360∘, 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 cot2θ−5cotθ+6=0\newlineAnswer: θ= \theta= θ=

Find all angles, $0^{\circ} \leq \theta<360^{\circ}$ , that satisfy the equation below, to the nearest tenth of a degree. $\newline$ $\cot ^{2} \theta-5 \cot \theta+6=0$ $\newline$ Answer: $\theta=$