How do you find the maximum values of $f\left( x \right) = 2\sin x + \cos x$ ?

Hint: In the given question, we are provided with a trigonometric expression involving the trigonometric functions sine and cosine and we are required to find the maximum value of the expression. We first divide such expressions by $\sqrt {{a^2} + {b^2}} $ to calculate the range and then transform the sum of two trigonometric functions into a single term. Then we will use the compound angle formula of cosine and sine to transform the expression into one single trigonometric function.

Complete step by step answer:
So, the given function is $f\left( x \right) = 2\sin x + \cos x$. So, we have to find the range of the trigonometric expression $\left( {a\sin x + b\cos x} \right)$. We divide such a trigonometric expression by $\sqrt {{a^2} + {b^2}} $ to transform the sum of two trigonometric ratios into one term. So, we have,
$f\left( x \right) = 2\sin x + \cos x$

We divide and multiply the expression by $\sqrt {{2^2} + {1^2}} = \sqrt 5 $. Hence, we get,
$ \Rightarrow f\left( x \right) = \sqrt 5 \left( {\dfrac{2}{{\sqrt 5 }}\sin x + \dfrac{1}{{\sqrt 5 }}\cos x} \right)$
Now, we can assume $\dfrac{2}{{\sqrt 5 }}$ as a cosine of some angle, say y.
Then, $\dfrac{1}{{\sqrt 5 }}$ will correspond to the sine of same angle y as we know that ${\sin ^2}x + {\cos ^2}x = 1$ and ${\left( {\dfrac{2}{{\sqrt 5 }}} \right)^2} + {\left( {\dfrac{1}{{\sqrt 5 }}} \right)^2} = 1$.

Hence, we have, $\sin y = \dfrac{1}{{\sqrt 5 }}$ and $\cos y = \dfrac{2}{{\sqrt 5 }}$.
Substituting the values, we get,
$ \Rightarrow f\left( x \right) = \sqrt 5 \left( {\cos y\sin x + \sin y\cos x} \right)$
Now, we know the compound angle formula for sine as $\left( {\cos \theta \sin \phi + \sin \theta \cos \phi } \right) = \sin \left( {\phi + \theta } \right)$. Hence, we get,
$ \Rightarrow f\left( x \right) = \sqrt 5 \sin \left( {x + y} \right)$
Now, we know that sine of any angle has a range of $\left[ { - 1,1} \right]$. So, the maximum value of sine function can be one. Hence, the maximum range of $\sqrt 5 \sin \left( {x + y} \right)$ is $\sqrt 5 $.

Therefore, the maximum value of $f\left( x \right) = 2\sin x + \cos x$ is $\sqrt 5 $.

Note: Such questions require grip over the concepts of trigonometry and inequalities. One must know the methodology to calculate the range of trigonometric expressions of the form $\left( {a\sin x + b\cos x} \right)$ in order to solve the given problem. Dividing or multiplying any inequality by a positive number does not change the signs of the inequality. But when we multiply or divide any inequality by a negative number, the signs of the inequality are reversed. Whereas, in the case of an equation, both sides remain equal if multiplied or divided by a positive or negative number.