On page 12 it's written that the function ($\delta>0$) $t(x)=1+\cos(x-x_0)-\cos(\delta)$ satisfies the following: $t(x)\ge 1$ in $I$ where $I = (x_0-\delta,x_0+\delta)$ $t(x)>1$ where $I'$ is some interior interval to $I$, and $|t(x)| \le 1$ outside of $I$.
My question how is it possible that there would be equality $t(x)=1$ in $I$ I mean it doesn't include the endpoints of the interval, I mean $t(x= x_0+\delta)=t(x_0-\delta)=1$ but $I$ doesn't include the endpoints of the interval.
Perhaps I should add also that $I\subset (-\pi , \pi)$.
When $|x-x_0|<\delta\le \pi$, we have
$$t(x)=1+\cos(x-x_0)-\cos(\delta)>1$$ by the monotonicity of cosine. Apparently, the author does not need strict inequality here: $t(x)\ge 1$ is enough for his purpose.
Writing $t(x)\ge 1$ does not mean that the equality is attained: it means not bothering to think whether it may be attained. This is a matter of writing style. Some mathematicians prefer using $\le $ and $\ge$ as "default mode" of inequalities, reserving $<$ and $>$ only for the situations when strictness matters. Benefits include not changing the inequality when passing to limits, or multiplying by $|x|$, etc.