Answer:
[tex]x = 0 , \pi , 2\pi[/tex]
Step-by-step explanation:
The given equation is:l
[tex] \tan^{2} (x) \sec^{2} x + 2 \sec^{2} x - \tan^{2} x = 2[/tex]
Add -2 to both sides of the equation to get:
[tex] \tan^{2} (x) \sec^{2} x + 2 \sec^{2} x - \tan^{2} x - 2 = 0[/tex]
We factor the LHS by grouping.
[tex]\sec^{2} x(\tan^{2} (x) + 2 ) - 1( \tan^{2} x + 2) = 0[/tex]
[tex](\sec^{2} x - 1)(\tan^{2} (x) + 2)= 0[/tex]
We now apply the zero product property to get:
[tex](\sec^{2} x - 1) = 0 \: \: or \: \: (\tan^{2} (x) + 2)= 0[/tex]
This implies that:
[tex]\sec^{2} x = 1 \: \: or \: \: \tan^{2} (x) = - 2[/tex]
[tex] \tan^{2} (x) = - 2 \implies \tan(x) = \pm \sqrt{ - 2} [/tex]
This factor is never equal to zero and has no real solution.
[tex]\sec^{2} x = 1[/tex]
This implies that:
[tex]\sec \: x= \pm\sqrt{1} [/tex]
[tex] \sec(x) = \pm - 1[/tex]
Recall that
[tex] \frac{1}{ \sec(x) } = \cos(x) [/tex]
We reciprocate both sides to get:
[tex] \cos(x) = \pm1[/tex]
[tex]\cos x = 1 \: or \: \cos x = - 1[/tex]
[tex]\cos x = 1 \implies \: x = 0 \: or \: 2\pi[/tex]
[tex]\cos x = - 1 \implies \: x = \pi[/tex]
Therefore on the interval
[tex]0 \leqslant x \leqslant 2\pi[/tex]
[tex]x = 0 , \pi , 2\pi[/tex]
