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Solve this trigonometric equation. $\frac{1}{\sqrt{2}}\left(\mathrm{sin}\left(\theta \right)+\mathrm{cos}\left(\theta \right)\right)=\frac{1}{\sqrt{2}}$
Divide both sides by $\frac{1}{\sqrt{2}}$
$\mathrm{sin}\left(\theta \right)+\mathrm{cos}\left(\theta \right)=1$
Divide both sides by $\mathrm{cos}\left(\theta \right)$
tan(θ)+1=sec(θ)
square both sides:
$\mathrm{tan}\left(\theta \right)+1=\mathrm{sec}\left(\theta \right)$
${\mathrm{tan}}^{2}\left(\theta \right)+2\mathrm{tan}\left(\theta \right)+1={\mathrm{sec}}^{2}\left(\theta \right)$
Use the identity ${\mathrm{sec}}^{2}\left(\theta \right)={\mathrm{tan}}^{2}\left(\theta \right)+1$:
${\mathrm{tan}}^{2}\left(\theta \right)+2\mathrm{tan}\left(\theta \right)+1={\mathrm{tan}}^{2}\left(\theta \right)+1$
$\therefore$
$2\mathrm{tan}\left(\theta \right)=0$

$\mathrm{tan}\left(\theta \right)=0$

$\theta =0,\pi ,2\pi$
I know that 0 and $2\pi$ are correct but that $\pi$ is wrong. I also know that the other correct answer is $\frac{\pi }{2}$.
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Freddy Doyle
The reason you got the extraneous solution $\theta =\pi$ is because you squared both sides of the equation $\mathrm{tan}\theta +1=\mathrm{sec}\theta$. You can check this by noting that $\mathrm{tan}\pi +1=1$ while $\mathrm{sec}\pi =-1$, so $\theta =\pi$ is a solution to the squared equation but not the original. On the other hand, you missed out the solution $\theta =\pi /2$ because you divided by $\mathrm{cos}\theta$ throughout, in which you've implicitly assumed $\mathrm{cos}\theta \ne 0$ and hence $\theta \ne \pi /2$
But these are rather easy to fix: check for extraneous solutions by substituting everything back into the original equation, and discuss the case $\mathrm{cos}\theta =0$ (i.e. $\theta =\pi /2$) separately. Other than these two issues, your solution is perfect (and quite smart, actually).