Solve $\cos (x)=-0.03$ on $0 \leq x< 2 \pi$

There are two solutions, A and B, with $A< B$
\[
\begin{array}{l}
A= \\
B=
\end{array}
\]

Step 1 ：We are given the equation \(\cos (x)=-0.03\) and we need to find the values of x that satisfy this equation in the interval \(0 \leq x<2 \pi\).

Step 2 ：Since the cosine function is periodic with period \(2\pi\), we know that there will be two solutions in this interval.

Step 3 ：The first step is to find the principal value, which is the value of x in the interval \(0 \leq x< \pi\) that satisfies the equation. We can do this by using the arccosine function, which is the inverse of the cosine function.

Step 4 ：The second step is to find the second solution. Since the cosine function is symmetric about the y-axis, the second solution will be the negative of the principal value, reflected about the line \(x=\pi\).

Step 5 ：By calculating these values, we find that \(A= 1.6008008286183735\) and \(B= 4.682384478561213\).

Step 6 ：\(\boxed{A= 1.6008008286183735, B= 4.682384478561213}\) are the solutions to the equation \(\cos (x)=-0.03\) on \(0 \leq x<2 \pi\).