# Solve this trigonometric equation. $\frac{1}{\sqrt2}(\sin(\theta)+\cos(\theta))=\frac{1}{\sqrt2}$

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?

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).

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.

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

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.

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