Question

find all solutions of each equation on the interval 0 less than or equal to x less than 2pi

tan^2x sec^2x +2sec^2x-tan^2x=2​

Answer 1

Answer:

$x=0,\pi,2\pi$

Step-by-step explanation:

The given trigonometric equation is

$\tan^2(x) \sec^2(x)+2\sec^2(x)-\tan^2(x)=2$.

Let us equate everything to zero

$\tan^2(x) \sec^2(x)+2\sec^2(x)-\tan^2(x)-2=0$.

We factor now to get:

$\sec^2(x)(\tan^2(x)+2)-1(\tan^2(x)+2)=0$.

$(\sec^2(x)-1)(\tan^2(x)+2)=0$.

Apply zero product property:

$(\sec^2(x)-1)=0,(\tan^2(x)+2)=0$.

$\sec^2(x)=1,\tan^2(x)=-2$.

When

$\sec^2(x)=1$.

$\implies \sec(x)=\pm1$.

Reciprocate both sides to get;

$\implies \cos(x)=\pm1$.

$x=0,\pi,2\pi$

When  $\tan^2(x)=-2$ we don't have real solutions.

Therefore $x=0,\pi,2\pi$ on $0\le x\le 2\pi$.