trigonometry - What have i done wrong in solving the general solution to $sec2theta=csc2theta$?

$$\sec2\theta=\csc2\theta$$

My attempt:

$$\begin{align}
\cos2\theta &= \sin2\theta \tag{1}\\
\cos^2\theta+\sin^2\theta-2\cos\theta\sin\theta &=0 \tag{2}\\
(\cos\theta-\sin\theta)^2 &=0 \tag{3}\\
\cos\theta-\sin\theta &=0 \tag{4}\\
\cos\theta &=\sin\theta \tag{5}\\
\tan\theta &=1 \tag{6}\\
\theta &=180^\circ n+45^\circ \quad\text{??} \tag{7}
\end{align}$$

But the answer was $90^\circ n+22.5^\circ$ and I'm not sure why. I've searched up the question online, and someone has proposed a solution where it is not factored; instead, the equation turns into $\tan2\theta=1$ on line $(2)$, and this allows you to get the correct solution.

What's wrong with factoring it though?

Answer

You have committed an error in the very beginning. Kindly note that

\begin{eqnarray}
\cos(2\theta)=\cos^{2}(\theta)-\sin^{2}(\theta).
\end{eqnarray}

You have written $\cos(2\theta)=\cos^{2}(\theta)+\sin^{2}(\theta)$, which is wrong.