Generally for such questions, creating a function out of the equation and solving it with different tools is much easier.
Clearly, $f(-5)>0$ and $f(1)<0$. Since, $f(x)$ is continuous function, $\exists c \in R$ such that $f(c)=0$.
So, we got one value for which $f(x)=0$. Now, we need to check if there are any other values for which $f(x)=0$.
We need to see if the function oscillates and takes multiple values of $x$ such that $f(x)=0$. Let’s start with the basic test of differentiation, to check if function is increasing or decreasing or oscillating, in $R$.
Now, $f’(x)=sin(2x)-3<0$ for all $x \in R$. Thus, $f(x)$ is a decreasing function in $R$.
Since, $f(x)$ is a decreasing function in $R$. So, we only have one point $c \in R$, for which $f(c)=0$.
Thus, $f(x)=0$ $\implies$ $sin^2(x)-3x-5=0$ $\implies$ $sin^2(x)-3x=5$, only for one point in $R$.
Ans: Number of points=1