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To solve the equation [tex]\(2 \sec^2 x - \tan^4 x = -1\)[/tex], we can use trigonometric identities and algebraic manipulation. Let's look at each step in detail:
1. Use the identity [tex]\(\sec^2 x = 1 + \tan^2 x\)[/tex]:
Replace [tex]\(\sec^2 x\)[/tex] with [tex]\(1 + \tan^2 x\)[/tex]:
[tex]\[ 2(1 + \tan^2 x) - \tan^4 x = -1 \][/tex]
2. Simplify the equation:
Distribute the 2 and bring all terms to one side:
[tex]\[ 2 + 2 \tan^2 x - \tan^4 x = -1 \][/tex]
[tex]\[ \tan^4 x - 2 \tan^2 x - 3 = 0 \][/tex]
3. Substitute [tex]\(u = \tan^2 x\)[/tex]:
This changes the equation into a quadratic form:
[tex]\[ u^2 - 2u - 3 = 0 \][/tex]
4. Solve the quadratic equation:
Using the quadratic formula [tex]\(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ u = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2} \][/tex]
This gives us two solutions:
[tex]\[ u_1 = \frac{2 + 4}{2} = 3 \][/tex]
[tex]\[ u_2 = \frac{2 - 4}{2} = -1 \quad (\text{not valid since } u = \tan^2 x \geq 0) \][/tex]
So we have:
[tex]\[ \tan^2 x = 3 \][/tex]
5. Find [tex]\(\tan x\)[/tex]:
Solving for [tex]\(\tan x\)[/tex]:
[tex]\[ \tan x = \pm \sqrt{3} \][/tex]
6. Determine the angles:
The values [tex]\(\tan x = \sqrt{3}\)[/tex] and [tex]\(\tan x = -\sqrt{3}\)[/tex] correspond to the angles:
[tex]\[ x = \frac{\pi}{3} + k\pi \quad \text{and} \quad x = \frac{2\pi}{3} + k\pi \][/tex]
These give us the general solutions in radians:
[tex]\[ x = \frac{\pi}{3} + k\pi \][/tex]
[tex]\[ x = \frac{2\pi}{3} + k\pi \][/tex]
[tex]\[ x = \frac{4\pi}{3} + k\pi \][/tex]
[tex]\[ x = \frac{5\pi}{3} + k\pi \][/tex]
Hence, the exact solutions are:
[tex]\[ x = \frac{\pi}{3} + k\pi, \quad \frac{2\pi}{3} + k\pi, \quad \frac{4\pi}{3} + k\pi, \quad \frac{5\pi}{3} + k\pi \][/tex]
So the correct answer is:
[tex]\[ \boxed{\frac{\pi}{3}+k \pi, \frac{2 \pi}{3}+k \pi, \frac{4 \pi}{3}+k \pi, \frac{5 \pi}{3}+k \pi} \][/tex]
1. Use the identity [tex]\(\sec^2 x = 1 + \tan^2 x\)[/tex]:
Replace [tex]\(\sec^2 x\)[/tex] with [tex]\(1 + \tan^2 x\)[/tex]:
[tex]\[ 2(1 + \tan^2 x) - \tan^4 x = -1 \][/tex]
2. Simplify the equation:
Distribute the 2 and bring all terms to one side:
[tex]\[ 2 + 2 \tan^2 x - \tan^4 x = -1 \][/tex]
[tex]\[ \tan^4 x - 2 \tan^2 x - 3 = 0 \][/tex]
3. Substitute [tex]\(u = \tan^2 x\)[/tex]:
This changes the equation into a quadratic form:
[tex]\[ u^2 - 2u - 3 = 0 \][/tex]
4. Solve the quadratic equation:
Using the quadratic formula [tex]\(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ u = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2} \][/tex]
This gives us two solutions:
[tex]\[ u_1 = \frac{2 + 4}{2} = 3 \][/tex]
[tex]\[ u_2 = \frac{2 - 4}{2} = -1 \quad (\text{not valid since } u = \tan^2 x \geq 0) \][/tex]
So we have:
[tex]\[ \tan^2 x = 3 \][/tex]
5. Find [tex]\(\tan x\)[/tex]:
Solving for [tex]\(\tan x\)[/tex]:
[tex]\[ \tan x = \pm \sqrt{3} \][/tex]
6. Determine the angles:
The values [tex]\(\tan x = \sqrt{3}\)[/tex] and [tex]\(\tan x = -\sqrt{3}\)[/tex] correspond to the angles:
[tex]\[ x = \frac{\pi}{3} + k\pi \quad \text{and} \quad x = \frac{2\pi}{3} + k\pi \][/tex]
These give us the general solutions in radians:
[tex]\[ x = \frac{\pi}{3} + k\pi \][/tex]
[tex]\[ x = \frac{2\pi}{3} + k\pi \][/tex]
[tex]\[ x = \frac{4\pi}{3} + k\pi \][/tex]
[tex]\[ x = \frac{5\pi}{3} + k\pi \][/tex]
Hence, the exact solutions are:
[tex]\[ x = \frac{\pi}{3} + k\pi, \quad \frac{2\pi}{3} + k\pi, \quad \frac{4\pi}{3} + k\pi, \quad \frac{5\pi}{3} + k\pi \][/tex]
So the correct answer is:
[tex]\[ \boxed{\frac{\pi}{3}+k \pi, \frac{2 \pi}{3}+k \pi, \frac{4 \pi}{3}+k \pi, \frac{5 \pi}{3}+k \pi} \][/tex]
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