Question
Find all solutions of 4tan^2θ−3=tan^4θ
on [0,2π).
θ=π/4, 3π/4, 5π/4, 7π/4
θ=π/3, 2π/3, 4π/3, 5π/3
θ=π/3, 2π/3, 4π/3, 5π/3, π/4, 3π/4,5π/4, 7π/4
θ=π/3, 4π/3, π/4, 3π/4
There are no solutions to this equation.
on [0,2π).
θ=π/4, 3π/4, 5π/4, 7π/4
θ=π/3, 2π/3, 4π/3, 5π/3
θ=π/3, 2π/3, 4π/3, 5π/3, π/4, 3π/4,5π/4, 7π/4
θ=π/3, 4π/3, π/4, 3π/4
There are no solutions to this equation.
Answers
Answered by
GPT-4o mini
To solve the equation \( 4\tan^2 \theta - 3 = \tan^4 \theta \), we can first rearrange it into a standard polynomial form. Let's rewrite the equation as:
\[
\tan^4 \theta - 4\tan^2 \theta + 3 = 0.
\]
Next, we will substitute \( x = \tan^2 \theta \). This transforms the equation into a quadratic equation:
\[
x^2 - 4x + 3 = 0.
\]
Now we can factor the quadratic equation:
\[
(x - 1)(x - 3) = 0.
\]
This gives us the solutions:
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1,
\]
\[
x - 3 = 0 \quad \Rightarrow \quad x = 3.
\]
Recalling that \( x = \tan^2 \theta \), we can revert back to \(\theta\):
1. For \( \tan^2 \theta = 1 \):
\[
\tan \theta = \pm 1.
\]
Thus, the possible values of \( \theta \) in the interval \( [0, 2\pi) \) are:
\[
\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}.
\]
2. For \( \tan^2 \theta = 3 \):
\[
\tan \theta = \pm \sqrt{3}.
\]
Thus, the possible values of \( \theta \) in the interval \( [0, 2\pi) \) are:
\[
\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}.
\]
Combining all the solutions together, we have:
\[
\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}.
\]
Therefore, the final answer is:
\[
\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}.
\]
The answer choice that represents this solution is:
\(\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}.\)
\[
\tan^4 \theta - 4\tan^2 \theta + 3 = 0.
\]
Next, we will substitute \( x = \tan^2 \theta \). This transforms the equation into a quadratic equation:
\[
x^2 - 4x + 3 = 0.
\]
Now we can factor the quadratic equation:
\[
(x - 1)(x - 3) = 0.
\]
This gives us the solutions:
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1,
\]
\[
x - 3 = 0 \quad \Rightarrow \quad x = 3.
\]
Recalling that \( x = \tan^2 \theta \), we can revert back to \(\theta\):
1. For \( \tan^2 \theta = 1 \):
\[
\tan \theta = \pm 1.
\]
Thus, the possible values of \( \theta \) in the interval \( [0, 2\pi) \) are:
\[
\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}.
\]
2. For \( \tan^2 \theta = 3 \):
\[
\tan \theta = \pm \sqrt{3}.
\]
Thus, the possible values of \( \theta \) in the interval \( [0, 2\pi) \) are:
\[
\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}.
\]
Combining all the solutions together, we have:
\[
\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}.
\]
Therefore, the final answer is:
\[
\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}.
\]
The answer choice that represents this solution is:
\(\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}.\)