Vrftsjtryk

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 $0 leqslant x < 2pi$

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 $sin x$:

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

$$5 cos^2 x = 1$$

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

$$cos x = ±sqrt{frac 1 5}$$

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

That gave us two answers within the range requested.

But let's replace $cos x$ instead:

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

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

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

$$4 = 5 sin^2 x$$

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

$$±sqrt frac 4 5 = sin x$$

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

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 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 $pmdfrac1{sqrt5}$, 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 $pmdfrac2{sqrt5}$, that will give yo only the solutions that belong to the domain of $arcsin$, which is $left[-dfracpi2,dfracpi2right]$. So, you will have to provide the extra solutions for your self. For instance, if you used the $arcsin$ function and you get a $alphainleft[-dfracpi2,0right)$, then use $2pi+alpha$ 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.

Notice that:

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

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.

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:

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

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

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

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

I guess you can take it from here