Solving Trig Equations Using Identities Exercise 3

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 \(x=\frac{\pi}{6},\frac{3\pi}{4}, \frac{5\pi}{6},\frac{7\pi}{6},\frac{7\pi}{4},\frac{11\pi}{6}\).