Find all solutions of each equation on the interval

Ответ:

florochoa217

02.01.2022

Mathematics

$x_1=0\\ \\x_2=\pi$

Step-by-step explanation:

Use definitions of $\tan x$ and $\sec x:$

$\tan x=\dfrac{\sin x}{\cos x}\\ \\\sec x=\dfrac{1}{\cos x}$

Hence, the equation is

$\left(\dfrac{\sin x}{\cos x}\right)^2\cdot \left(\dfrac{1}{\cos x}\right)^2+2\left(\dfrac{1}{\cos x}\right)^2-\left(\dfrac{\sin x}{\cos x}\right)^2=2$

Multiply this equation by $\cos ^4x:$

$\sin^2 x+2\cos^2 x-\sin^2x\cos^2x=2\cos^4x\\ \\\sin^2 x+2\cos^2 x-\sin^2x\cos^2x-2\cos^4x=0\\ \\\sin^2x(1-\cos^2x)+2\cos^2x(1-\cos^2x)=0\\ \\(1-\cos^2x)(\sin^2x+2\cos^2x)=0$

This means that

$1-cos^2x=0\ \text{or} \ \sin^2x+2\cos^2x=0$

The second equation has no solutions, because the sum of two non-negative numbers is 0 only if both these numbers are 0 and this is impossible, because $\sin x$ and $\cos x$ cannot be 0 simultaneously.