Question

find the range of 1 upon (cos 2x+2sin2x​)

Answer 1

Given:

$\bold{\frac{1}{(\cos 2x + 2\sin 2x)}}$

To prove:

Find range.

Solution:

$\Rightarrow \frac{1}{(\cos 2x + 2\sin 2x)}$

Let the function is f(x):

$f(x)= {(\cos 2x + 2 \sin 2x)}$

derivative the above method:

$f'(x)= {(- 2 \sin(2x) + 2 \times 2 \cos (2x))}\\\\f'(x)= {(- 2 \sin(2x) + 4\cos (2x))}$

let the method is equal to 0:

$0 = {- 2 \sin(2x) + 4 \cos (2x))}\\\\ \sin(2x) = 2 cos (2x) \\\\\tan 2x = 2\\\\x= \frac{1}{2} \tan^{-1} (2)$

putting the value of x is above method:

$\cos^{-1} \theta = (\frac{B}{H}) = \frac{1}{\sqrt{5}} \\\\\sin^{-1} \theta = (\frac{P}{H}) = \frac{2}{\sqrt{5}}$

$f(\frac{1}{2} \tan^{-1}) = cos(\tan^{-1} (2)) + 2 \sin (\tan^{-1} (2))$

                 $= cos(\tan^{-1} (2)) + 2 \sin (\tan^{-1} (2)) \\\\= cos(\cos^{-1} (\frac{1}{\sqrt{5}})) + 2 \sin (\sin^{-1} \frac{2}{\sqrt{5}}) \\\\= \frac{1}{\sqrt{5}} + \frac{4}{\sqrt{5}} \\\\= \frac{1+4}{\sqrt{5}} \\\\= \frac{5}{\sqrt{5}} \\\\= {\sqrt{5}}$

Range:

$- \sqrt{5} \leq f(x) \leq \sqrt{5}$

The equation is given in the form of 1 upon that's why value is:

$\bold { \Rightarrow - \frac{1}{\sqrt{5}} \geq \frac{1}{f(x)} \geq \frac{1}{\sqrt{5}}}$

let the method is g(x): $g(x) \geq \frac{1}{\sqrt{5}} , \ \ \ \ g(x) \leq - \frac{1}{\sqrt{5}}$

The range is:  $\bold{g(x) \geq \frac{1}{\sqrt{5}} , \ \ \ \ g(x) \leq - \frac{1}{\sqrt{5}}}$