We solve \(\sin z + \cos z = 1\) for \(z \in [0^\circ, 360^\circ]\).

February 22, 2026

$We solve \(\sin z + \cos z = 1\) for \(z \in [0^\circ, 360^\circ]\).$

Related Articles

The dot product is always zero, regardless of \(x\). Hence, the vectors are orthogonal for all \(x\).
Thus, the solution is all real numbers \(x\), and the answer can be expressed as \(\boxed{\text{All } x \in \mathbb{R}}\).
Find the sum of all angles \(z \in [0^\circ, 360^\circ]\) that satisfy \(\sin z + \cos z = 1\).
Square both sides:
+ 2\sin z \cos z = 1 \Rightarrow 2\sin z \cos z = 0 \Rightarrow \sin 2z = 0
So, \(2z = 0^\circ, 180^\circ, 360^\circ, 540^\circ, \ldots\), giving:

Trending Articles