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

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?

• The easiest thing to do is just plug all answers you find back into the original identity and keep the ones for which it's true! – user113102 Dec 6 '18 at 19:08
• A quick method would be to observe from the equation $\sec x = 2 \csc x$ that both $\sin x, \cos x$ must have the same sign, so $x$ must lie in the first or third quadrant. $2.03, -1.1$ get easily rejected because they lie in the second and fourth quadrants respectively. – Shubham Johri Dec 6 '18 at 19:42

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.

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.

• That I forgot this is a true indication it's been a long time since a math class! Of course you always plug your answers back into the original! – rrauenza Dec 6 '18 at 19:09

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.

• Did you mean to say $2|\cos x|=|\sin x|$? – Shubham Johri Dec 6 '18 at 19:40

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