An application of the intermediate value theorem Show that between any two real solutions of $e^x \sin x = 1 $, there is at least one real solution of $e^x \cos x = -1 $
Try: Let $f(x) = e^x \sin x - 1 $.
Consider $I = [0, \pi/2]$, Then, $f( 0) = -1 $ and $f( \pi/2) = e^{\pi/2} - 1 >0 $. Thus, one can find a solution $c_1 \in I $ such that $f(c_1) = 0$. Similarly, if $J= [ \pi/2, \pi ] $, one can find $c_2 \in J$ such that $f(c_2) = 0 $. Consider the interval $[c_1,c_2]$. Let $F(x) = e^x \cos x + 1 $. We have
$$F(c_1 ) = e^{c_1} \cos (c_1) + 1 < e^{c_1} \cos 0 +1 =  e^{c_1} + 1 >0 $$
and
$$ F(c_2) = e^{c_2} \cos (c_2) - 1 < e^{c_2} \cos \pi - 1 = - e^{c_2} - 1 < 0 $$
so, $\mathbf{there \; exists}$ some $\alpha \in [c_1,c_2]$ such that $F(\alpha) = 0$ as we wanted to prove.
Is this a correct proof?
 A: No, this is not a correct proof: you only found a root for the second equation between two specific roots of the first equation, not between any two roots. I actually doubt it can be done by the Intermediate Value Theorem.
But you can apply Rolle's Theorem here. Assume $c_1,c_2$ are two roots of the first equation. Verify that your function $f(x)=e^x\sin x-1$ satisfies the conditions of Rolle's Theorem on the interval $[c_1,c_2]$. Then the conclusion of Rolle's Theorem – that there exists at least one point $c\in(c_1,c_2)$ such that $f'(c)=0$ – will give you the desired result. (Note: subtracting $1$ is fine but optional, as it works just fine with $f(x)=e^x\sin x$.)
UPDATE. The above solution was wrong (shame on me!), as the OP pointed out in the comment. But it still can be done using Rolle's Theorem with one extra trick before applying it. Since $e^x\neq0$ for any real $x$, we can safely divide by it. So we can transform the originally given equation:
$$e^x\sin x=1 \Leftrightarrow \sin x =\frac{1}{e^x} \Leftrightarrow \sin x=e^{-x} \Leftrightarrow \sin x-e^{-x}=0.$$
In other words, any solution to $e^x\sin x=1$ is a solution to $\sin x-e^{-x}=0$ and vice versa. Now let $f(x)=\sin x-e^{-x}$, and proceed as above, i.e. apply Rolle's Theorem on $[c_1,c_2]$, which are any two roots of the (either) equation. You'll get a root of $f'(x)=\cos x+e^{-x}$, which can be transformed back into the desired result.
