Solving Trig Equations Using Identities Exercise 4

Solution: \(x=0,\dfrac{\pi}{6}, \pi, \dfrac{11\pi}{6},2\pi\)

Solution Method: We want to turn this equation into something we can factor. Thinking about the identities I can apply to this equation, the ratio identity for tangent comes to mind. So let’s try substituting \(\tan{x}=\frac{\sin{x}}{\cos{x}}\). \[\begin{aligned} \sqrt{3}\tan{x}&=2\sin{x} \\ \sqrt{3}\left(\frac{\sin{x}}{\cos{x}}\right)&=2\sin{x} \end{aligned}\] Then I multiply both sides by cosine, and move everything to one side. \[\begin{aligned} \sqrt{3}\sin{x} = 2\sin{x}\cos{x} \\ \sqrt{3}\sin{x} - 2\sin{x}\cos{x} = 0 \end{aligned}\] This equation I can factor. I can factor out a \(\sin{x}\). \[\begin{aligned} \sin{x}(\sqrt{3} - 2\cos{x}) = 0 \end{aligned}\]So now I have two factors I can set equal to 0.

First, \[\sin{x} = 0\] This is a quadrantal value and the values satsifying sine equal to 0 are \(x=0,\pi,2\pi\). (We need to include both 0 and \(2\pi\) because both are included in the interval)

Second, \[\begin{aligned} \sqrt{3}-2\cos{x} &= 0 \\ \sqrt{3} &= 2\cos{x} \\ \frac{\sqrt{3}}{2} &= \cos{x} \end{aligned}\] The reference angle satisfying \(\cos{x}=\frac{\sqrt{3}}{2}\) is \(\frac{\pi}{6}\). And cosine is positive in Q1 and Q4. So the values in \([0,2\pi]\) satisfying the equation are \(x=\frac{\pi}{6}\) in Q1, and \(x=\frac{11\pi}{6}\) in Q4.

So all the values satisfying \(\sqrt{3}\tan{x}=2\sin{x}\) in \([0,2\pi]\) are \(x=0,\frac{\pi}{6}, \pi, \frac{11\pi}{6}, 2\pi\).