# What is wrong with this way of solving trig equations?

Let's suppose I have to find the values of $\theta$ and $\alpha$ that satisfy these equations:

• $\cos^3 \theta$ = $\cos \theta$
• $3\tan^3 \alpha = \tan \alpha$

on the interval $[0; 2 \pi]$.

If I try to solve, for instance, the first equation like this:

$$\cos^3 \theta = \cos \theta$$

$$\cos^2 \theta * \cos \theta = \cos \theta$$

$$\cos^2 \theta = \cos \theta \div \cos \theta$$

$$\cos^2 = 1$$

$$\cos \theta = \pm \sqrt {1}$$

$$\cos \theta = \pm 1$$

I end up getting only:

$\theta = 0 ; \pi ; 2 \pi$

But I know that $\pi /2$ and $3\pi /2$ would also make the equation true since $\cos^3 (\pi /2) = cos (\pi /2) = 0$ and $\cos^3 ( 3 \pi /2) = cos (3 \pi /2) = 0$.

The same problem arises when I try to solve the second equation:

$$3\tan^3 \alpha = \tan \alpha$$ $$3\tan^2 \alpha * \tan \alpha = \tan \alpha$$ $$3\tan^2 \alpha = \tan \alpha \div \tan \alpha$$ $$3\tan^2 \alpha =1$$ $$\tan^2 \alpha =1/3$$ $$\tan \alpha =\pm \sqrt{1/3}$$ $$\tan \alpha =\pm 1/\sqrt{3}$$

The values of $\alpha$ that make the tangent equal to $\pm 1/\sqrt{3}$ between 0 and $2 \pi$ are only these:

$\alpha = \pi / 6 ; 5 \pi /6 ; 7 \pi / 6 ; 11 \pi /6$

However, I should also find $0$ and $\pi$ because $3\tan^3 (0) = \tan (0) = 3 tan^3 (\pi) = tan ( \pi ) = 0$.

Something similar happens when I'm looking for the local minima and maxima of this function: $$f(x) = \sin^2(x) + \cos(x)$$

on the interval $[0; 2 \pi]$

$f'(x) = 2 \sin(x) \cos(x) - \sin(x)$

$0 = 2 \sin(x) \cos(x) - \sin(x)$

$\sin(x) = 2 \sin(x) \cos(x)$

$\sin(x) / \sin(x) = 2 \cos(x)$

$1 = 2 \cos(x)$

$1/2 = \cos(x)$

$x = \pi / 3 ; 5 \pi /3$

And again, plugging $x = 0$ or $x= \pi$ or $x = 2 \pi$ also make the derivative 0.

I've noticed that in all of these 3 cases I have the same trigonometric function on both sides of the equation and I'm dividing both sides by that function. This is making one side of the equation equal to 1 at some point. What is wrong or incomplete with this method? Why am I missing some results when I do this?

• You forgot that $A^3\iff A\equiv A^2=1$ OR $A=0$. Mar 30 '17 at 22:55
• In the first case, you divided-though by $\cos\theta$, recklessly ignoring the possibility that $\cos\theta$ could itself be zero (which it is, at precisely the angles you note are missing from your solution set). Likewise when dividing-through by $\tan\alpha$.
– Blue
Mar 30 '17 at 22:57

The equation $x^3=x$ has three solutions: you can write it as $$x^3-x=0$$ so $$x(x-1)(x+1)=0$$ and the roots are $0$, $1$ and $-1$.

You cannot “divide by $x$”, which is the mistake you make when you “divide by $\cos\theta$”.

Thus your equation becomes $$\cos\theta=0 \quad\text{or}\quad \cos\theta=1 \quad\text{or}\quad \cos\theta=-1$$

The second equation is similar: $$\tan\alpha=0 \quad\text{or}\quad \tan\alpha=\frac{1}{\sqrt{3}} \quad\text{or}\quad \tan\alpha=-\frac{1}{\sqrt{3}}$$

Clearly, when $$\cos \theta = 0$$, then $$\cos^3 \theta = \cos \theta$$.

So, in the interval $$[0, 2\pi]$$, we get $$\theta \in \left\{\frac 12 \pi, \frac 32 \pi \right\}$$

If $$\cos \theta \ne 0$$, then we can divide both sides by $$\cos \theta$$...