Answer
Use the Intermediate Value Theorem
Work Step by Step
Let $f(x)$ = $3x+2\cos x+5$
Then
$f(0)$ = $7$ $\gt$ $0$
$f(-\pi)$ = $-3\pi+3<0$
Since $f$ is continuous on $R$, the Intermediate Value Theorem assures us that there is at least one zero of $f$ in $[-\pi,0]$.
$f'(x)$ = $3-2\sin x$
Because $\sin x\in[-1,1]$ it follows that $f(x)\in[1,5]$.
$f'(x)$ $\gt$ $0$ implies that $f$ is increasing on $[-\pi,0]$, so there is exactly one zero of $f$, therefore exactly one real root of the equation $3x+2\cos x+5$ = $0$ in that interval.
