Question

The number of solutions of $2\sin x + \cos x = 3$ is
1. $1$
2. $2$
3. Infinite
4. No solution

Answer 1

Hint: There is a trigonometric function given in the question. First, let us see what is the range of that function. This is going to give us an idea about where the solution of the function might exist. Later, we can compare with the given number and figure out if it has a solution or not. If it has a solution, we need to calculate that as well.

Complete step-by-step solution:
Consider the given trigonometric equation $2\sin x + \cos x = 3$. The LHS can be taken to find out the range of the equation.
We know that the range of an equation of the form $a\cos x + b\sin x$ is $ - \sqrt {\left( {{a^2} + {b^2}} \right)} \leqslant a\cos x + b\sin x \leqslant \sqrt {\left( {{a^2} + {b^2}} \right)} $
Now let us consider values for $a$ and $b$.
Let, $a = 1,b = 2$
Substituting these values in the above equation, we get
$\eqalign{
  & \Rightarrow - \sqrt {\left( {{1^2} + {2^2}} \right)} \leqslant a\cos x + b\sin x \leqslant \sqrt {\left( {{1^2} + {2^2}} \right)} \cr
  & \Rightarrow - \sqrt 5 \leqslant a\cos x + b\sin x \leqslant \sqrt 5 \cr} $
Now, we know the range of the equation, $a\cos x + b\sin x$.
The solution lies between $ - \sqrt 5 $and $\sqrt 5 $.
But, the given equation is $2\sin x + \cos x = 3$.
The RHS is $3$ which is greater than $\sqrt 5 $
Since the range does not match the given equation, there is no solution for the given equation.
Hence, option (4) is the correct answer.

Note: Remember the formula for range of equations for all the combinations. To find their values, substitute small numbers, thereby finding out the range of the equation. While comparing, be careful about the roots or negative integers, because they will be lesser than real numbers.