Equations which contains polynomial and trigonometric functions do not show explicit solutions and numerical methods should be used to find the roots.
The simplest root finding method is Newton; starting from a reasonable guess $x_0$, the method will update it according to $$x_{n+1}=x_n-\frac{F(x_n)}{F'(x_n)}$$ So, in your case, $$F(x)=x \cos(x)-\sin(x)$$ I think it is better to let it under this form because of the discontinuities of $\tan(x)$.
You can notice that if $x=a$ is a root, $x=-a$ will be another root. So, let us just focus on $0\leq x \leq 3\pi$. If you plot the function, you notice that, beside the trivial $x=0$, there are two roots located close to $5$ and $8$. These would be the guesses.
Using $F'(x)=-x \sin (x)$, the iterative scheme then write $$x_{n+1}=x_n-\frac{1}{x_n}+\cot (x_n)$$ Let us start with $x_0=5$; the method then produces the following iterates : $4.50419$, $4.49343$, $4.49341$ which is the solution for six significant figures.
I let you doing the work for the other solution.
You have correctly found the critical points, that is, the points at which the first derivative is equal to zero. To determine the relative extrema, we have two options. We can apply the First Derivative Test or the Second Derivative Test.
First Derivative Test. Let $f$ be a function that is continuous on $[a, b]$ and differentiable on $(a, b)$ except possibly at a point $c$.
(a) If $f'(x) > 0$ for all $x < c$ and $f'(x) < 0$ for all $x > c$, then $f$ has a relative maximum at $c$.
(b) If $f'(x) < 0$ for all $x < c$ and $f'(x) > 0$ for all $x > c$, then $f$ has a relative minimum at $c$.
Second Derivative Test. If $f$ has a critical point $c$ and $f''$ exists in an open interval $(a, b)$ and
(a) if $f''$ is negative in $(a, b)$, then $f$ has a relative maximum at $c$;
(b) if $f''$ is positive in $(a, b)$, then $f$ has a relative minimum at $c$.
In this case, it is easier to apply the Second Derivative Test. We have
\begin{align*}
f(x) & = x - 2\cos x\\
f'(x) & = 1 + 2\sin x\\
f''(x) & = 2\cos x
\end{align*}
As you determined, the critical points are
\begin{align*}
x & = \frac{7\pi}{6} + 2k\pi, k \in \mathbb{Z}\\
x & = \frac{11\pi}{6} + 2k\pi, k \in \mathbb{Z}
\end{align*}
At the points $x = \frac{7\pi}{6} + 2k\pi, k \in \mathbb{Z}$, which lie in the third quadrant, $f''(x) = \cos x < 0$, so the Second Derivative Test tells us these points are relative maxima. At the points $x = \frac{11\pi}{6} + 2k\pi, k \in \mathbb{Z}$, which lie in the fourth quadrant, $f''(x) = \cos x > 0$, so the Second Derivative Test tells us these points are relative minima.
We get the same result if we apply the First Derivative Test. Since the sine function decreases from $0$ to $-1$ in the third quadrant, the first derivative
$f'(x) = 1 + 2\sin x$ changes from positive to negative at the points $x = \frac{7\pi}{6} + 2k\pi, k \in \mathbb{Z}$, so the First Derivative Test tells us that these points are relative maxima. Since the sine function increases from $-1$ to $0$ in the fourth quadrant, the first derivative $f'(x) = 1 + 2\sin x$ changes from negative to positive at the points $x = \frac{11\pi}{6} + 2k\pi, k \in \mathbb{Z}$, so the First Derivative Test tells us that these points are relative minima.
Best Answer
$\cos x$ ranges from $-1$ to $1$, and $\sin x$ is increasing on $(-\frac{\pi}{2},\frac{\pi}{2})$, so since $[-1,1]\subset (-\frac{\pi}{2},\frac{\pi}{2})$ the minimum and maximum are $\sin(-1)$ and $\sin(1)$, respectively.