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I am trying to solve $\sqrt{3}\tan\theta=2\sin\theta$ on the interval $[-\pi,\pi]$.

$$\sqrt{3}\tan\theta=2\sin\theta \Rightarrow \sqrt{3}=\frac{2\sin\theta}{\tan\theta}$$

$$\Rightarrow \sqrt{3}=2\sin\theta \cdot \frac{\cos\theta}{\sin\theta} \Rightarrow 3 = 4\cos^2\theta$$

I get $\displaystyle \cos \theta = \pm{\frac{\sqrt{3}}{2}}$; the cosine of $30^{\circ}$ and $150^{\circ}$ so arrived at the solutions $\displaystyle -\frac{5}{6}\pi,-\frac{1}{6}\pi,\frac{1}{6}\pi,\frac{5}{6}\pi$.

Looking in the back of the book (and checking with Wolfram), the answer is $\displaystyle -\pi,-\frac{1}{6}\pi,0,\frac{1}{6}\pi,\pi$.

Where am I going wrong please?

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+1 for showing work, so asking a specific question. – Ross Millikan May 19 '11 at 13:49
+1 for the same reason. – Américo Tavares May 19 '11 at 14:06
In the future please write \sin, \cos, \tan instead of just sin, cos, tan. – Américo Tavares May 19 '11 at 14:13
up vote 4 down vote accepted

Why square the equation? $\cos \theta = \frac{\sqrt 3}{2}$ is easier to solve. Also, include the solutions of $\sin \theta = 0$ before canceling $\sin \theta$.

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The squaring introduced the spurious roots. – Ross Millikan May 19 '11 at 13:47
@Ross, that was also my point. – lhf May 19 '11 at 13:50
I was typing up an answer as yours came it. It seemed better to make that explicit as one could think the squaring just made more work. – Ross Millikan May 19 '11 at 13:53
Yes, the squaring was completely pointless. Can't believe I did that. I need to learn to think laterally. Thanks for pointing it out. – PeteUK May 19 '11 at 14:09
@Ross You're right - I didn't get from the reading of this answer that the spurious roots were introduced by the squaring. – PeteUK May 19 '11 at 14:10

When you square both sides of a equation you get a new equation that has all the solutions of the first but may admit other solutions. That was in the step $\sqrt{3}=2\cos\theta \Rightarrow 3=4\cos^2 \theta$.

Added: Also you can only divide your initial equation by $\sin \theta\neq 0$. But for $\theta=0$ or $\theta=\pm\pi$, $\sin\theta=\tan\theta =0$, which are solutions of your initial equation.

Added 2: In general the equation $A^n=B^n$ has all the solutions of the equation $A=B$ but may have additional ones. This is a consequence of the algebraic identity $$A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+\cdots +AB^{n-2}+B^{n-1}).$$

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érico: thank you for cleaning up my initial post, and your great answer. With regards to being careful about the division, this is something I'm going to have to keep an eye on. It feels like a subtle/difficult topic for some reason... – PeteUK May 19 '11 at 14:22
@PUK: You are welcome. – Américo Tavares May 19 '11 at 14:26

Dear PUK, you missed $\theta=0$ because you divided the equation by $\tan\theta$ in the first step which is only possible if $\tan\theta$ is nonzero and finite. That's why $\theta=0$ and $\theta=\pi$ and $\theta=-\pi$ for which $\tan\theta$ vanish must be separately checked and indeed, you will find out that the equation is satisfied because $0=0$.

On the other hand, you added the wrong solutions $\pm 5\pi/6$ because their squared cosine is $3/4$, but the cosine itself has a wrong sign, so your squaring created problems. The squaring was unnecessary, as pointed out by lhf as I was writing this sentence, but if you still want to square it, you have to check all the solutions that they work and you will find out that the $\pm 5\pi/6$ solutions don't.

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Thanks for your answer. I see the squaring caused me even more problems. – PeteUK May 19 '11 at 14:14

When you cancelled the $\sin \theta$ factors, you lost some solutions. This is similar to starting with $x^2 = x$, cancelling an $x$ and concluding that $x=1$; the obvious, $x=0$ solution has been lost. The remedy for this is to factor instead. In the example I provided, that would amount to writing $x^2 - x = 0$, then factoring to get $x(x-1)=0$ then setting each factor equal to zero, as usual.

You also gained solutions when you squared. This is similar to starting with $x=1$, then squaring to get $x^2 = 1$, so $x= \pm 1$. Notice that in your problem, squaring was superfluous since a few steps later you took a square root that "undid" that squaring anyway.

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Appreciate the clear response. – PeteUK May 19 '11 at 14:24

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