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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?

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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).

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All your steps are correct.

However, when you divided by $\cos \theta$, you implicitly assumed that $\cos \theta \ne 0$. Therefore, you should add the possible solutions $\theta = \frac{\pi}{2}, \frac{3 \pi}{2}$. One way to remind yourself of this is to write:

$$\tan \theta + 1 = \sec \theta \tag{$\cos \theta \ne 0$}$$

In addition, when you squared both sides, you also introduced possible extraneous solutions. As a result, when you have all the possible solutions: $0, \frac{\pi}{2}, \pi, \frac{3 \pi}{2}, 2 \pi$, you need to substitute all of them into the original equation.

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Hint: $$ \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4} = \cos \frac{\pi}{4}, $$ so $$ \frac{1}{\sqrt2}(\sin(\theta))+\cos(\theta)) = \sin \frac{\pi}{4} \sin(\theta) + \cos \frac{\pi}{4} \cos \theta = \cos (\theta - \frac{\pi}{4}). $$

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By manipulation you have added some more solutions which need to be excluded.

More simply from here by squaring both sides

$$\sin(\theta)+\cos(\theta)=1 \implies 2\cos \theta \sin \theta =\sin (2\theta)=0$$

that is $2\theta=k\pi\implies \theta=k\frac \pi 2$, then check for the solutions which satisfies the original equation.

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  • $\begingroup$ +1 Good point that the OP does not have to divide both sides by $\cos \theta$. $\endgroup$
    – Toby Mak
    Oct 26, 2019 at 10:20

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