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

`x_1=0\\ \\x_2=\pi`

Explanation:

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

`\tan x=(\sin x)/(\cos x)\\ \\\sec  x=(1)/(\cos x)`

Hence, the equation is

`\left((\sin x)/(\cos x)\right)^2\cdot \left((1)/(\cos x)\right)^2+2\left((1)/(\cos x)\right)^2-\left((\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`