Solve $\cos^2x-\sin^2x= 1$ I am trying to solve for $x$ in the following equation:
$$\cos^2x-\sin^2x= 1$$  
Given that $\cos^2x+\sin^2x= 1$, is this something I could use to solve it?
 A: Subtract two equation you get $2(\sin^2x)=0 \implies \sin^2x=0 \implies \sin x=0 \implies x = \pi\cdot n, n\in \mathtt{Z} $.
A: Observe that $\cos^2 x$ never gets bigger than 1.  So if there is a solution, it must have $\cos^2 x = 1$ and $\sin^2 x = 0$. Does that help?
A: I set up the problem like this:
$\cos^{2}(x) - \sin^2(x) = 1 \rightarrow$  I used the half angle rule,
$\frac{1}{2}\cdot(1+\cos(2x)) - \frac{1}{2}(1- \cos(2x)) = 1$
$\frac{1}{2}\cdot((1 + (\cos(2x) - 1 + \cos(2x)) = 1$
$(2\cdot \cos(2x)) = 2$
$\cos(2x) = 1$
$x = 0$ , $\pi$, $2\pi$, and so on . . .  $x =\pi n$.
A: If you don't know the identity $ \cos{2x} =\cos^2x-\sin^2x$, you can often solve the problem by expressing $\sin$ and $\cos$ through the exponential function:
$$\left(\frac{e^{ix}+e^{-ix}}{2}\right)^2 - \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^2 = 1$$
$$ \frac{e^{2ix}+ 2e^{-2ix}+e^{-2ix}}{4} + \frac{e^{2ix} - 2e^{-2ix}+e^{-2ix}}{4} =1$$
$$ \frac{e^{2ix}+e^{-2ix}}{2} = 1 $$
$$ \cos(2x) = 1 $$
A: Use the identity $\cos 2x =\cos^2x-\sin^2x$.
A: You could indeed use that, since then $$\cos^2x-\sin^2x=\cos^2x+\sin^2x.$$ Gather all your trig terms on one side, and the rest is almost trivial.
A: Given
$$\cos^2x-\sin^2x= 1\tag1$$
Known
$$\cos^2x+\sin^2x= 1\tag2$$
$(1)\quad+\quad(2)$
$$\Rightarrow 2\cos^2x= 2$$
$$\Rightarrow \cos^2x= 1$$
$$\Rightarrow \cos x= \pm1$$ 
$$x = n\pi$$ 
A: $$\cos^2x-\sin^2x= 1$$
$$1-\sin^2x-\sin^2x= 1$$
$$-2\sin^2x=0\iff\sin^2x=0\iff\sin x=0\iff x=k\pi$$  
A: Maximum value of $\cos^2(x)$ = $1$
Minimum value of $\sin^2(x)$ = $0$
So this is the only case where you get $\cos^2(x) - \sin^2(x) = 1$.
Hence $\cos^2(x) = 1$ and $\sin^2(x) = 0$ => $x = nπ$
Or you could have used the formula : $\cos^2(x) - \sin^2(x) = \cos(2x)$
Hope the answer is clear !
