# What have i done wrong in solving the general solution to $\sec2\theta=\csc2\theta$?

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

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?

• Second line, $\cos 2\theta = \cos^2\theta - \sin^2 \theta$. – Arthur Aug 3 '16 at 10:41
• Noooooo. Ok thankyou – kjhg Aug 3 '16 at 10:41
• BTW why don't you solve is as $\tan 2\theta=1$? – David Quinn Aug 3 '16 at 11:12

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