I tried solving this equation as follows where $0\leq\theta\leq2\pi$:
$$\frac{1}{\sqrt2}(\sin(\theta)+\cos(\theta))=\frac{1}{\sqrt2}$$
Divide both sides by $\frac{1}{\sqrt2}$.
$$\sin(\theta)+\cos(\theta)=1$$
Divide both sides by $\cos(\theta)$.
$$\tan(\theta)+1=\sec(\theta)$$
square both sides:
$$(\tan(\theta)+1)^2=\sec^2(\theta)$$
$$\tan^2(\theta)+2\tan(\theta)+1=\sec^2(\theta)$$
Use the identity $\sec^2(\theta)=\tan^2(\theta)+1$:
$$\tan^2(\theta)+2\tan(\theta)+1=\tan^2(\theta)+1$$
$\therefore$
$$2\tan(\theta)=0$$
$\therefore$
$$\tan(\theta)=0$$
$\therefore$
$$\theta=0,\pi,2\pi$$
I know that 0 and $2\pi$ are correct but that $\pi$ is wrong. I also know that the other correct answer is $\frac{\pi}{2}$.
Where did I go wrong?
Best Answer
The reason you got the extraneous solution $\theta=\pi$ is because you squared both sides of the equation $\tan\theta+1=\sec\theta$. You can check this by noting that $\tan\pi+1=1$ while $\sec\pi=-1$, 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 $\cos\theta$ throughout, in which you've implicitly assumed $\cos\theta\neq0$ and hence $\theta\neq\pi/2$.
But these are rather easy to fix: check for extraneous solutions by substituting everything back into the original equation, and discuss the case $\cos\theta=0$ (i.e. $\theta=\pi/2$) separately. Other than these two issues, your solution is perfect (and quite smart, actually).