Different answers with $\sec(x) = 2\csc(x)$

Question

My son and I were solving this last night and we get different answers depending on which identities we use. The question also did specify $0 \leqslant x < 2\pi$

Here's our work:

$$\sec x = 2 \csc x$$

$$\frac 1 {\cos x} = \frac 2 {\sin x}$$

cross multiply:

$$2 \cos x = \sin x$$

and square both sides (I think this introduces a problem?)

$$4 \cos^2 x = \sin^2 x$$

Now we used the identity $\sin^2 x + \cos^2 x = 1$

Let's replace $\sin x$:

$$4 \cos^2 x = 1 - \cos^2 x$$

$$5 \cos^2 x = 1$$

$$\cos^2 x = \frac 1 5$$

$$\cos x = ±\sqrt{\frac 1 5}$$

$$\cos^{-1}\left(±\sqrt \frac 1 5\right) = 1.10, 2.03$$

That gave us two answers within the range requested.

But let's replace $\cos x$ instead:

$$4 \cos^2 x = \sin^2 x$$

$$4 (1 - \sin^2 x) = \sin^2 x$$

$$4 - 4 \sin^2 x = \sin^2 x$$

$$4 = 5 \sin^2 x$$

$$\frac 4 5 = \sin^2 x$$

$$±\sqrt \frac 4 5 = \sin x$$

$$\sin^{-1}\left(±\sqrt \frac 4 5\right) = x = 1.1, -1.1$$

Two answers, but we can throw out the negative one because it is not within the range specified.

Then we used the $\tan x$ identity (which is what we should have done to begin with since squaring obviously seems to introduce invalid answers):

$$\tan x = \frac {\sin x} {\cos x}$$

$$2 \cos x = \sin x$$

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

$$2 = \tan x$$

$$\tan^{-1} 2 = 1.1$$

So now I assume $1.1$ is the right answer. But where did $-1.1$ and $2.03$ come from?

They don't show up in the graphs:

AH! But they do show up in the squared version, which I now understand is where the extra answers came from:

What is the fundamental mistake here? How would one use the squaring method, and then at the end know which solution(s) to throw out as a side effect?

Answer 1

The two first methods led to $\cos^2x=\frac15$ and to $\sin^2x=\frac45$. That's the same assertion, since $\cos^2x+\sin^2x=1$.

But if you apply the $\arccos$ function to $\pm\dfrac1{\sqrt5}$, that will give you only the solutions that belong to the domain of $\arccos$, which is $[0,\pi]$. And if you apply the $\arcsin$ function to $\pm\dfrac2{\sqrt5}$, that will give yo only the solutions that belong to the domain of $\arcsin$, which is $\left[-\dfrac\pi2,\dfrac\pi2\right]$. So, you will have to provide the extra solutions for your self. For instance, if you used the $\arcsin$ function and you get a $\alpha\in\left[-\dfrac\pi2,0\right)$, then use $2\pi+\alpha$ instead; it is also a solution and it belongs to the right range.

Finally, if you are solving an equation of the type $f(x)=g(x)$ and if $x_0$ is such that $f^2(x_0)=g^2(x_0)$, then what you have to do is to compute $f(x_0)$ and $g(x_0)$. Either they'r equal or they're symmetric. If they're equal, then you have a solution in your hands. Keep it. Otherwise, throw it away.

Answer 2

Squaring an equation can create extraneous solutions. For instance (as a trivial example), the equation $x=1$ has the solution $x=1$, but if we square it we get $x^2=1$ which has solutions $x=1,-1$. To check which "solutions" are indeed correct after solving by squaring, one can simply plug them back into the original equation: you throw out ones which do not solve the original equation. So for our example, we obtained $x=1,-1$ as "solutions" after squaring, but now we plug $x=-1$ back into the original equation and find that $1=-1$, so this is not a solution.

Answer 3

Notice that:

$$4\cos^2(x)=\sin^2(x) \to 2\cos(|x|)=\sin(|x|)$$

And the source of your problem becomes apparent.

While $\cos(|x|)=\cos (x)$, we have that $\sin(|x|)=-\sin(x)$ for $x<0$, and this explains why you get $-1.1$ as a solution here.

Answer 4

1) $a^2 = b^2$ => $a = b$ or $a = - b$

2) $a = b$ => (square both sides) $a^2 = b^2$

The idea here is that in the first case you have that $a = -b$, but in the second case (which is your case), your $a^2 = b^2$ inevitably adds the $a = -b$ solutions to your total, which obviously are wrong since your original equation is $a = b$

The proper way to solve this problem is to do this:

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

$\cos(x)*\frac{2}{\sqrt{1^2+2^2}} - \sin(x)*\frac{1}{\sqrt{1^2+2^2}} = 0$

$\cos(x)*\cos(\arccos(\frac{2}{\sqrt{5}})) - \sin(x)*\sin(\arcsin(\frac{1}{\sqrt{5}})) = 0$

$\cos(x + \arccos(\frac{2}{\sqrt{5}})) = 0$

I guess you can take it from here