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tags:
Fundamental Trigonometric Identity Sine Tangent Trigonometry Trigonometry Series

Solving Trig Equations Using Identities Exercise 4

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 values \(x \in [0,2\pi]\) satisfying  \(\sqrt{3}\tan{x}=2\sin{x}.\)

Video Errata: At 6:05, the presenter says and writes \(\frac{5\pi}{6},\) but it should be \(\frac{11\pi}{6}.\)

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 \(x=0,\frac{\pi}{6}, \pi, \frac{11\pi}{6}, 2\pi\).

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