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The Number of Points $x \in \Re$ for which $\sin ^{2} x-3x=5$ is ,

1. 0
2. 1
3. more than one but finite
4. $\infty$

Number of points $x \in \mathbb{R}$ will be $1.$

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, $\cos^2 x – \sin^2 x = \cos 2x \implies 1- 2\sin^2 x = \cos 2x$

$\sin^2 x = \frac{1}{2} – \frac{1}{2}\cos 2x$

Nice approach sir .

Sir can you suggest some source to practice more of these kind of problems.
don’t know bhai...sorry. probably previous year questions of JEE and ISI contain these type of questions.

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 $R$.

Ans: Number of points=1

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