# Thread: Solve cos^2(...)+cos^2(...)-2 = 0 ?

1. ## Solve cos^2(...)+cos^2(...)-2 = 0 ?

anyone have any clever ideas on how to solve this for x?

$cos^2(\pi*16^x)+cos^2(\pi*16^{1-x})-2 = 0$

?
$(cos(\pi*16^x)-1)(cos(\pi*16^x)+1)=-(cos(\pi*16^{1-x})-1)(cos(\pi*16^{1-x})+1)$
?

2. Originally Posted by pinion
anyone have any clever ideas on how to solve this for x?

$cos^2(\pi*16^x)+cos^2(\pi*16^{1-x})-2 = 0$

?
$(cos(\pi*16^x)-1)(cos(\pi*16^x)+1)=-(cos(\pi*16^{1-x})-1)(cos(\pi*16^{1-x})+1)$
?
Since $|\cos(x)| \le 1$ this equation implies that:

$\cos^2(\pi*16^x)=1$

and

$\cos^2(\pi*16^{1-x})=1$

RonL

3. Originally Posted by CaptainBlack
Since $|\cos(x)| \le 1$ this equation implies that:

$\cos^2(\pi*16^x)=1$

and

$\cos^2(\pi*16^{1-x})=1$

RonL
huh?

4. Originally Posted by pinion
huh?
Your equation asserts that the sum of two positive quataties that are both
less than or equal to 1 is 2, so both must be exactly 1.

RonL

PS huh? is uninformative, please explain yourself more clearly in future

5. Originally Posted by CaptainBlack
Since $|\cos(x)| \le 1$ this equation implies that:

$\cos^2(\pi*16^x)=1$

and

$\cos^2(\pi*16^{1-x})=1$

RonL
If you require further help in finding the solutions let us know, but the problem is trivial from here.

RonL

6. Thanks for being so nice about me not explaining in the manner you require, my bad. But as to the 'huh', I was firstly confused at how your statement followed from the equation, and secondly confused at why I would/should observe that fact to solve the equation. Is that the only way? Or do you suggest that to be the easiest way? I don't know. But I do know this follows from that observation -
$cos(\pi*16^x) = 1$
and
$cos(\pi*16^x) = -1$
and
$cos(\pi*16^{1-x}) = 1$
and
$cos(\pi*16^{1-x}) = -1$
and since the arccos of both 1 and -1 is $c*\pi$ (or more specifically $c_1*\pi$ and $c_2*\pi$ where $c_1$ is even and $c_2$ is odd) I would now have to find $c_1,c_2,c_3,c_4$ - so unless there is an algebraic method for doing that, splitting the equation in two doesn't help me. Is there anything else you can recommend?

7. Originally Posted by pinion
Thanks for being so nice about me not explaining in the manner you require, my bad. But as to the 'huh', I was firstly confused at how your statement followed from the equation, and secondly confused at why I would/should observe that fact to solve the equation. Is that the only way? Or do you suggest that to be the easiest way? I don't know. But I do know this follows from that observation -
$cos(\pi*16^x) = 1$
and
$cos(\pi*16^x) = -1$
and
$cos(\pi*16^{1-x}) = 1$
and
$cos(\pi*16^{1-x}) = -1$
and since the arccos of both 1 and -1 is $c*\pi$ (or more specifically $c_1*\pi$ and $c_2*\pi$ where $c_1$ is even and $c_2$ is odd) I would now have to find $c_1,c_2,c_3,c_4$ - so unless there is an algebraic method for doing that, splitting the equation in two doesn't help me. Is there anything else you can recommend?
That:

$
\cos^2(\pi*16^x)=1
$

implies that $\pi 16^x=k \pi$ for some integer $k$, which then implies that $16^x=2^{4x}=k$ but this implies that:

$4x=\log_2(k)/2$

or:

$x=\log_2(k)/8$.

Similarly (using the other equation): we find that for some integer n:

$1-x=\log_2(n)/8$.

So:

$\log_2(k)/8=1-\log_2(n)/8$

or:

$\log_2(k/n)=8$

so:

$k/n=2^8.$

so every integer $n$ determines a solution $x=1-\log_2(n)/8$.

(don't rely on the algebra above being right as I cannot check it in detail at present, either I or another helper will correct any mistakes when we get the chance).

RonL

8. Originally Posted by pinion
Thanks for being so nice about me not explaining in the manner you require, my bad. But as to the 'huh', I was firstly confused at how your statement followed from the equation, and secondly confused at why I would/should observe that fact to solve the equation. Is that the only way? Or do you suggest that to be the easiest way? I don't know.
The reason why the obsevation is worth making is that it is useful information
about the nature of the solutions that is otherwise not obvious.

Other approaches are likely to lead you through some horendous trig transformations that are unlikely to go anywhere.

RonL

9. thanks, i'll look into trig transformations.