Question

Find all solutions of each equation on the interval $0\leq x\ \textless \ 2\pi .$

$tan^{2} x$$sec^{2} x$ + 2 $sec^{2}x$$-tan^{2} x$ =2

Answer 1

Answer:

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

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.

Solve the first equation:

$\cos^2x=1\Rightarrow \\ \\\cos x=1\ \text{or}\ \cos x=-1\\ \\x=2\pi k,\ k\in Z\ \ \text{or}\ \ x=\pi+2\pi k,\ k\in Z$

If $0\le x<2\pi,$ then

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