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tags:
Cosecant Cotangent Fundamental Trigonometric Identity Trigonometry Trigonometry Series

Solving Trig Equations Using Identities Exercise 3

Author: Hannah Solomon

The following problem is solved in this video. It is recommended that you try to solve the problem before watching the video. You can click "Reveal Answer" to see the answer to the problem.

Problem: Find all \(0\leq x \leq 2\pi\) satisfying \(\cot^3{x}-3\cot{x}=4-\csc^2{x}.\)

Solution: \(x=\dfrac{\pi}{6},\dfrac{3\pi}{4},\dfrac{5\pi}{6},\dfrac{7\pi}{6},\dfrac{7\pi}{4},\dfrac{11\pi}{6}\)

Solution Method: This equation is a cubic with two different trig functions, cotangent and cosecant. I do not want to mess with this in terms of two variables, so that’s where the Pythagorean identity comes in. The Pythagorean identity \(1+\cot^2{x}=\csc^2{x}\) relates cotangent and cosecant. So I’m going to use it to substitute \(\csc^2{x}\) in terms of cotangent; because the rest of my terms are cotangent. So we have \[\begin{aligned} \cot^3{x}-3\cot{x}&=4-\csc^2{x} \\ \cot^3{x}-3\cot{x}&=4-(1+\cot^2{x}) \\ \cot^3{x}-3\cot{x}&=4-1-\cot^2{x} \\ \cot^3{x}-3\cot{x}&=3-\cot^2{x} \end{aligned}\] Now I’m going to set all the terms equal to zero. \[\begin{aligned} \cot^3{x} + \cot^2{x} -3\cot{x} -3 &= 0 \end{aligned}\] So now I have this cubic in cotangent. I want to be able to factor this, so let’s see if I can do it by grouping. I can take a \(\cot^2{x}\) out of the first two terms and a -3 out of the last two terms. \[\begin{aligned} \cot^2{x}(\cot{x}+1) -3(\cot{x} +1) &= 0 \end{aligned}\] The factors match so, indeed, I can factor the cubic by grouping as \[\begin{aligned} (\cot^2{x}-3)(\cot{x}+1) &= 0 \end{aligned}\] So now I can take these two binomials and set them equal to 0 and solve for x.

First, \[\begin{aligned} \cot^2{x} - 3 &= 0 \\ \cot^2{x} & = 3 \\ \sqrt{\cot^2{x}} & = \pm \sqrt{3} \\ \cot{x} &= \pm \sqrt{3} \end{aligned}\] So now I need to think about what reference angle produces cotangent of \(\sqrt{3}\). Remember, \(\cot{x}= \frac{\cos{x}}{\sin{x}}\). It is not 0 or \(\pm\)1, so it will not be a quadrantal value. At \(\frac{\pi}{4}\) sine and cosine are both \(\frac{\sqrt{2}}{2}\), so \(\cot{\frac{\pi}{4}}=\frac{\cos{\frac{\pi}{4}}}{\sin{\frac{\pi}{4}}}=1\). So that leaves me with \(\frac{\pi}{6}\) or \(\frac{\pi}{3}\). At \(\frac{\pi}{6}\), \(\cot{\frac{\pi}{6}}=\frac{\cos{\frac{\pi}{6}}}{\sin{\frac{\pi}{6}}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\), so \(\frac{\pi}{6}\) is the reference value satisfying \(\cot{x} = \sqrt{3}\). Now, the \(\pm\) means we are not restricting to any quadrants, so we include the \(\frac{\pi}{6}\) values for every quadrant. So the values in \([0,2\pi]\) satisfying the equation are \(x=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}\).

Now for the second factor, \[\begin{aligned} \cot{x} &=1 \\ \cot{x} &= -1 \end{aligned}\] Cotangent is equal to \(\pm\) 1 when sine and cosine are the same, which we know occurs at multiples of \(\frac{\pi}{4}\). And cotangent is negative (ASTC, cotangent matches sign of tangent) in Q2 and Q3. So the values satisfying the equation at \(x=\frac{3\pi}{4}, \frac{7\pi}{4}\).

So then the values of \(x \in [0,2\pi]\) satisfying \(\cos^3{x}-3\cot{x}=4-\csc^2{x}\) are \(x=\frac{\pi}{6},\frac{3\pi}{4}, \frac{5\pi}{6},\frac{7\pi}{6},\frac{7\pi}{4},\frac{11\pi}{6}\).

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