How do you solve for $x$ in $\cos \left( -100 \right)=\cos \left( 55 \right)\cos x+\sin \left( 55 \right)\sin x$?

Hint: We can easily contract the right hand side of the above equation as it is a trigonometric expansion. For this we need to use the trigonometric identity $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$, by which the RHS will get reduced to a cosine term. The LHS is also a cosine term, so the equation obtained can be solved by using the general solution of $\cos \theta =\cos \alpha $ which is given by $\theta =2n\pi \pm \alpha $, $n\in Z$.

Complete step by step solution:
The equation given in the above question is
$\cos \left( -100 \right)=\cos \left( 55 \right)\cos x+\sin \left( 55 \right)\sin x$
We can see that in the right hand side of the above equation, a trigonometric expansion is written. So we can contract it to simplify the RHS of the above equation by using the trigonometric identity given by
$\begin{align}
  & \Rightarrow \cos \left( A-B \right)=\cos A\cos B+\sin A\sin B \\
 & \Rightarrow \cos A\cos B+\sin A\sin B=\cos \left( A-B \right) \\
\end{align}$
Substituting $A=55$ and $B=x$ in the above identity, we get
\[\Rightarrow \cos \left( 55 \right)\cos x+\sin \left( 55 \right)\sin x=\cos \left( 55-x \right)\]
Substituting this in the given equation, we get
$\begin{align}
  & \Rightarrow \cos \left( -100 \right)=\cos \left( 55-x \right) \\
 & \Rightarrow \cos \left( 55-x \right)=\cos \left( -100 \right) \\
 & \Rightarrow \cos \left( -\left( x-55 \right) \right)=\cos \left( -100 \right) \\
\end{align}$
Now, we know that $\cos \left( -A \right)=\cos A$. So the above equation can also be written as
$\Rightarrow \cos \left( x-55 \right)=\cos \left( 100 \right)$
Now, we know that the general solution of the equation $\cos \theta =\cos \alpha $ is given by $\theta =2n\pi \pm \alpha $. Where $n\in Z$. This means that the solution of the above equation can be given by
$\Rightarrow x-55=2n\pi \pm 100$
Adding $55$ on both the sides, we finally get the solution as
$\begin{align}
  & \Rightarrow x-55+55=2n\pi \pm 100+55 \\
 & \Rightarrow x=2n\pi \pm 155 \\
\end{align}$
Hence, the solution of the given equation is $x=2n\pi \pm 155$, $n\in Z$.

Note:
We may be tempted to solve the equation $\cos \left( x-55 \right)=\cos \left( 100 \right)$, which is obtained in the above solution by equating the arguments of both the cosine terms. But that will give us only a single solution of the given equation. We must remember that since the trigonometric functions are periodic functions, infinitely many solutions of a trigonometric equation can exist. So the infinitely many solutions of a trigonometric equation are always written in terms of $n$, an arbitrary integer.