Find all solutions of each equation on the interval 0 ≤ x < 2 pi. tan^2 x sec^2 x +2 sec^2 x - tan^2 x=2 SOMEONE PLEASE HELPPP!!

Question

asked Mar 2, 2020 159k views

1 Answer

Jonathon Anderson · Answer 1 · 2020-03-08T00:28:21+0000

Answer:

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

Explanation:

The given equation is:l

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

Add -2 to both sides of the equation to get:

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

We factor the LHS by grouping.

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

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

We now apply the zero product property to get:

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

This implies that:

$\sec^(2) x = 1 \: \: or \: \: \tan^(2) (x) = - 2$

$\tan^(2) (x) = - 2 \implies \tan(x) = \pm √( - 2)$

This factor is never equal to zero and has no real solution.

$\sec^(2) x = 1$

This implies that:

$\sec \: x= \pm√(1)$

$\sec(x) = \pm - 1$

Recall that

$(1)/( \sec(x) ) = \cos(x)$

We reciprocate both sides to get:

$\cos(x) = \pm1$

$\cos x = 1 \: or \: \cos x = - 1$

$\cos x = 1 \implies \: x = 0 \: or \: 2\pi$

$\cos x = - 1 \implies \: x = \pi$

Therefore on the interval

$0 \leqslant x \leqslant 2\pi$

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