How do you solve $\cos (\theta ) - \sin (\theta ) = 1?$

Hint: Try to convert the equation in a form such that you can apply compound angle formula of trigonometric identities to the given trigonometric equation.
Since, this equation contains $\sin $ and $\cos $ functions only, so you should divide the equation (i.e. both sides) with $\sqrt {{{(coefficient\;of\;\sin \theta )}^2} + {{(coefficient\;of\;\cos \theta )}^2}} $ in order to get the required equation in which you can easily apply compound angle formula of trigonometric identities.

Complete answer:
The given trigonometric equation which has to be solved is
$\cos (\theta ) - \sin (\theta ) = 1$
Now, we should divide both the sides with $\sqrt {{{(coefficient\;of\;\sin \theta )}^2} + {{(coefficient\;of\;\cos \theta )}^2}} $
As from the equation we came to know that coefficient of $\sin \theta $ is $1$ and coefficient of $\cos \theta $ is also $1$
So the required divisor will be
$
   = \sqrt {{{(coefficient\;of\;\sin \theta )}^2} + {{(coefficient\;of\;\cos \theta )}^2}} \\
   = \sqrt {{1^2} + {1^2}} \\
   = \sqrt {1 + 1} \\
   = \sqrt 2 \\
 $
Now dividing both sides of the equation $\cos (\theta ) - \sin (\theta ) = 1$ with $\sqrt 2 $ , we will get
$
   \Rightarrow \dfrac{{\cos (\theta ) - \sin (\theta )}}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }} \\
   \Rightarrow \dfrac{{\cos (\theta )}}{{\sqrt 2 }} - \dfrac{{\sin (\theta )}}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }} \\
 $
Can we simplify this furthermore? Or write it all in a trigonometric equation?
Yes, we can write this, as we know that
$\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
So we can replace $\dfrac{1}{{\sqrt 2 }}$ in terms of $\sin \dfrac{\pi }{4}$ and $\cos \dfrac{\pi }{4}$
Now replacing and rewriting the equation as
$ \Rightarrow \cos (\theta ) \times \cos \dfrac{\pi }{4} - \sin (\theta ) \times \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
Are we familiar with this type of equation?
Actually yes, you should have seen this type of trigonometric equation before in the compound angle formula of trigonometric identities.
And this trigonometric equation is looking parallel to this trigonometric identity
$\cos x\cos y - \sin x\sin y = \cos (x + y)$
Isn’t it similar to our equation,
So let us continue solving our equation with help of the above trigonometric identity.
$
   \Rightarrow \cos (\theta ) \times \cos \dfrac{\pi }{4} - \sin (\theta ) \times \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} \\
   \Rightarrow \cos \left( {\theta + \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }} \\
 $
We know that the general solution of $\cos x = \dfrac{1}{{\sqrt 2 }}$ is
$x = 2n\pi \pm \dfrac{\pi }{4}$ , where $n \in I$
That means,
$ \Rightarrow \theta + \dfrac{\pi }{4} = 2n\pi \pm \dfrac{\pi }{4}$, where $n \in I$
$ \Rightarrow \theta = 2n\pi \pm \dfrac{\pi }{4} - \dfrac{\pi }{4}$, where $n \in I$
Therefore general solution of $\cos (\theta ) - \sin (\theta ) = 1$ is $\theta = 2n\pi \pm \dfrac{\pi }{4} - \dfrac{\pi }{4}$

Note: General equation of $\cos x = \dfrac{1}{{\sqrt 2 }}$ is $x = 2n\pi \pm \dfrac{\pi }{4}$ , where $n \in I$ but if you can’t find it in the options, then convert the terms into sine angles compound formula, i.e. $\sin x\cos y - \sin y\cos x = \sin (x - y)$.