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Pre-Calculus A Unit Test Unit Test Active 1 4 5 6 7 8 9 10 \square TIME REMAINING 03:31:58
Which shows all the exact solutions of 2sec2xtan4x=12 \sec ^{2} x-\tan ^{4} x=-1 ? Give your answer in radians. π3+kπ\frac{\pi}{3}+k \pi and 2π3+kπ\frac{2 \pi}{3}+k \pi π3+2kπ\frac{\pi}{3}+2 k \pi and 5π3+2kπ\frac{5 \pi}{3}+2 k \pi π4+2kπ,3π4+2kπ,5π4+2kπ\frac{\pi}{4}+2 k \pi, \frac{3 \pi}{4}+2 k \pi, \frac{5 \pi}{4}+2 k \pi, and 7π4+2kπ\frac{7 \pi}{4}+2 k \pi π3+kπ,2π3+kπ,4π3+kπ\frac{\pi}{3}+k \pi, \frac{2 \pi}{3}+k \pi, \frac{4 \pi}{3}+k \pi, and 5π3+kπ\frac{5 \pi}{3}+k \pi

Studdy Solution

STEP 1

What is this asking? We need to find all the angles xx, in radians, that make the equation 2sec2xtan4x=12 \sec ^{2} x-\tan ^{4} x=-1 true. Watch out! Remember those tricky trigonometric identities!
And don't forget about adding +2kπ+2k\pi or +kπ+k\pi to account for *all* the solutions.

STEP 2

1. Rewrite the equation
2. Simplify and solve
3. Find all solutions

STEP 3

Let's **rewrite** everything in terms of tanx\tan x.
Remember the super helpful identity sec2x=1+tan2x\sec^2 x = 1 + \tan^2 x!
Substituting this into our equation gives us 2(1+tan2x)tan4x=12(1+\tan^2 x) - \tan^4 x = -1.

STEP 4

**Distribute** the 2 to get 2+2tan2xtan4x=12 + 2\tan^2 x - \tan^4 x = -1.

STEP 5

Now, let's **move everything** to one side to get a nice, neat zero on the other: tan4x2tan2x3=0\tan^4 x - 2\tan^2 x - 3 = 0.
Look! This is like a quadratic equation, but with tan2x\tan^2 x instead of just xx.
Think of it as (tan2x)22(tan2x)3=0(\tan^2 x)^2 - 2(\tan^2 x) - 3 = 0.

STEP 6

Let's **factor** this bad boy!
We're looking for two numbers that multiply to -3 and add up to -2.
That's -3 and 1!
So, we have (tan2x3)(tan2x+1)=0(\tan^2 x - 3)(\tan^2 x + 1) = 0.

STEP 7

Now, we can **set each factor equal to zero** and solve!

STEP 8

First, tan2x3=0\tan^2 x - 3 = 0.
This means tan2x=3\tan^2 x = 3, so tanx=±3\tan x = \pm\sqrt{3}.

STEP 9

Second, tan2x+1=0\tan^2 x + 1 = 0.
This means tan2x=1\tan^2 x = -1.
Uh oh!
We can't take the square root of a negative number, so there are no solutions from this part.

STEP 10

Okay, so we're left with tanx=3\tan x = \sqrt{3} and tanx=3\tan x = -\sqrt{3}.

STEP 11

For tanx=3\tan x = \sqrt{3}, our **principal solutions** are x=π3x = \frac{\pi}{3} and x=4π3x = \frac{4\pi}{3}.
Since the tangent function has a period of π\pi, all solutions are of the form π3+kπ\frac{\pi}{3} + k\pi, where kk is an integer.

STEP 12

For tanx=3\tan x = -\sqrt{3}, our **principal solutions** are x=2π3x = \frac{2\pi}{3} and x=5π3x = \frac{5\pi}{3}.
Again, because the tangent function has a period of π\pi, all solutions are of the form 2π3+kπ\frac{2\pi}{3} + k\pi, where kk is an integer.

STEP 13

Our solutions are x=π3+kπx = \frac{\pi}{3} + k\pi and x=2π3+kπx = \frac{2\pi}{3} + k\pi.

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