Find the general solution of $\sqrt2 \sec\theta + \tan\theta = 1$
What I have tried:
$$\frac{\sqrt2}{\cos\theta} + \frac{\sin\theta}{\cos\theta} = 1$$
$$\sqrt2 +\sin\theta = \cos\theta$$
$$\cos\theta - \sin\theta = \sqrt2$$
$$\cos (\frac{\pi}{4} + \theta) =1$$
$$\implies \theta = 2n\pi - \frac{\pi}{4}, n\in \mathbb{Z}$$
However, from the equation $\cos (\frac{\pi}{4} + \theta) =1$ can be represented as $\sin (\frac{\pi}{4} - \theta) =1$ which when calculated becomes $\theta = \frac{\pi}{4} - m\pi + (-1)^{m+1} \frac{\pi}{2}, m\in \mathbb{Z}$.
Substituting $m= 1,2,3,...$ & $n=1,2,3,..$, I get a different set of solutions for both cases. May I know why? or Where have I gone wrong?
THANK YOU!
$\sin y=1\implies y=2m\pi+\dfrac\pi2$
If $2n\pi-\dfrac\pi4=-2m\pi-\dfrac\pi2+\dfrac\pi4$
$n=-m$