If $f(x) = \sin^2x$ and $g(x) = 3x + 5$ then for $f(x)=g(x),$ answer depends on, for how many points $f(x)$ and $g(x)$ cut each other.

We can plot line $3x+5$ easily and since $\sin^2 x$ is an even function, so it will be symmetric about Y-axis and also, it is non-negative function, so, you have to make graph of $\sin^2 x$ in first quadrant and then reflect it about Y-axis.

Since, $-1 \leq \sin x \leq +1 \implies 0 \leq \sin^2 x \leq +1,$ So, if we are going from $\sin x$ to $\sin^2 x$ and so, values get squared and range becomes $[0,1].$

To do systematically, we can write, $\cos^2 x – \sin^2 x = \cos 2x \implies 1- 2\sin^2 x = \cos 2x $

Generally for such questions, creating a function out of the equation and solving it with different tools is much easier.

Consider, $f(x)=sin^2(x)-3x-5$.

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 $R$.