Solve tan(theta) = 1 for all real theta.

Step-by-step solution

1. 1

   Find the reference angle from the definition. Since $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$, a tangent of $1$ means $\sin\theta=\cos\theta$. In a right triangle that says the opposite and adjacent legs are equal — the $45^\circ$-$45^\circ$-$90^\circ$ triangle. So

   $\theta=45^\circ=\frac{\pi}{4}$

   is a solution, confirmed exactly by $\sin45^\circ=\cos45^\circ=\tfrac{\sqrt2}{2}$.

2. 2

   Locate the second solution in a full turn. Tangent is also positive in the third quadrant, where sine and cosine are both negative and still equal: at $\theta=\tfrac{5\pi}{4}$, $\sin=\cos=-\tfrac{\sqrt2}{2}$, so $\tan=1$ again.

3. 3

   Recognise the period is pi, not 2pi. The gap between those two solutions is exactly $\pi$. That is because negating both sine and cosine leaves their ratio unchanged: $\tan(\theta+\pi)=\tan\theta$. So tangent repeats every half turn — unlike sine and cosine, which need $2\pi$.

4. 4

   Write the general solution as a single family. Because the period is $\pi$, both base solutions are captured at once:

   $\theta=\frac{\pi}{4}+k\pi,\qquad k\in\mathbb{Z}$

   Taking $k=0$ gives $\tfrac{\pi}{4}$ and $k=1$ gives $\tfrac{5\pi}{4}$. No second family is needed here — a genuine difference from $\sin x=c$ or $\cos x=c$, which each require two.

5. 5

   Give the degree form.

   $\theta=45^\circ+180^\circ k,\qquad k\in\mathbb{Z}$

   so $45^\circ,\ 225^\circ,\ 405^\circ,\ -135^\circ,\ \ldots$

6. 6

   Verify across several k. Evaluating $\tan\left(\tfrac{\pi}{4}+k\pi\right)$ for $k=-3,\,0,\,1,\,5$ gives $1.000000$ in every case ✓, and none of these angles lands on an odd multiple of $\tfrac{\pi}{2}$ where tangent would be undefined ✓.