George was tasked with solving the equation $$4\cdot\cos\left(\frac52x+\frac{\pi}{8}\right)=-\sqrt8$$ for real values of $x$. Let us follow his solution:
(1) After dividing the equation by $4$ and simplifying the right side, he obtained the equation in the form: $$\cos\left(\frac52x+\frac{\pi}{8}\right)=-\frac{\sqrt2}{2}$$
(2) Using the substitution $\left(\frac52x+\frac{\pi}{8}\right)=a$, he transformed the equation into: $$\cosa=-\frac{\sqrt2}{2}$$
(3) George solved the above trigonometric equation for the substituted variable $a$ and found the solutions: $$a_1=\frac34\pi+k\cdot2\pi,\quad a_2=\frac54\pi+k\cdot2\pi,$$ where $k\in\mathbb{Z}$.
(4) He then substituted these values of $a$ back into the substitution and obtained two linear equations for $x$: $$\frac52x_1+\frac{\pi}{8}=\frac34\pi,\quad\frac52x_2+\frac{\pi}{8}=\frac54\pi$$
(5) Finally, he expressed $x_1$ and $x_2$ from the above equations: $$x_1=\frac14\pi+k\cdot2\pi,\quad x_2=\frac{9}{20}\pi+k\cdot2\pi,$$ where $k\in\mathbb{Z}$.
However, George made a mistake in one of his steps. Identify this step.
The mistake is in step (1). He made a mistake when simplifying the right side of the equation. The equation should have been simplified to: $$\cos\left(\frac52x+\frac{\pi}{8}\right)=-\frac{\sqrt2}{8}$$
The mistake is in step (2). The substitution should have been $\frac52x=a$. The correct equation for $a$ would then be: $$\cos a=-\frac{\sqrt2}{2}-\frac{\pi}{8}$$
The mistake is in step (3). There is no need to express two solutions (using the period). The correct solution to the substituted equation is: $$a=\frac34\pi+k\cdot2\pi,\quad\mbox{where}\ k\in\mathbb{Z}$$
The mistake is in step (4). It's necessary to substitute the whole solutions for $a$, including the multiples of the period, back into the substitution. Afterward, the smallest period for $x$ should be $\frac45\pi$.
The mistake is in step (5). The unknown variable $x$ was incorrectly expressed from both equations. George should have obtained: $$x_1=\frac{25}{16}\pi+k\cdot\frac15\pi,\quad x_2=\frac{45}{16}\pi+k\cdot\frac15 \pi,\quad k\in\mathbb{Z}$$
In steps (1)-(3), George proceeded correctly. He obtained: $$a_1=\frac34\pi+k\cdot2\pi,\quad a_2=\frac54\pi+k\cdot2\pi,$$ where $k\in\mathbb{Z}$. Next, we must substitute these values (including the period) back into the substitution: $$\frac52x_1+\frac{\pi}{8}=\frac34\pi+k\cdot2\pi,\quad\frac52x_2+\frac{\pi}{8}=\frac54\pi+k\cdot2\pi$$ We then solve for $x$ from these linear equations, and the solutions to the given initial equation are: $$x_1=\frac14\pi+k\cdot\frac45\pi,\quad x_2=\frac{9}{20}\pi+k\cdot\frac45\pi,$$ where $k\in\mathbb{Z}$.