Question
cos theta = 1 + 2 sec theta
Solve the following equation for all values on the interval below:
0 is less than or equal to theta, theta is less than 2pi
Solve the following equation for all values on the interval below:
0 is less than or equal to theta, theta is less than 2pi
Answers
GPT-4o mini
To solve the equation \( \cos \theta = 1 + 2 \sec \theta \), we can start by rewriting \( \sec \theta \) in terms of \( \cos \theta \):
\[
\sec \theta = \frac{1}{\cos \theta}
\]
Substituting this into the equation gives:
\[
\cos \theta = 1 + \frac{2}{\cos \theta}
\]
Next, we can multiply both sides of the equation by \( \cos \theta \) (noting that \( \cos \theta \neq 0 \), so we must check separately if \( \cos \theta = 0 \)):
\[
\cos^2 \theta = \cos \theta + 2
\]
Rearranging gives us a quadratic equation:
\[
\cos^2 \theta - \cos \theta - 2 = 0
\]
This can be factored as follows:
\[
(\cos \theta - 2)(\cos \theta + 1) = 0
\]
Setting each factor to zero gives:
1. \( \cos \theta - 2 = 0 \) which gives \( \cos \theta = 2 \) (not possible since cosine values are always between -1 and 1).
2. \( \cos \theta + 1 = 0 \) which gives \( \cos \theta = -1 \).
Next, we solve for \( \theta \) when \( \cos \theta = -1 \):
\[
\theta = \pi
\]
Now, we need to check if there are any angles where \( \cos \theta = 0 \) that would make our original assumption invalid. The values of \( \theta \) where \( \cos \theta = 0 \) are:
\[
\theta = \frac{\pi}{2}, \frac{3\pi}{2}
\]
However, substituting these values in the original equation:
1. For \( \theta = \frac{\pi}{2} \):
\[
\cos\left(\frac{\pi}{2}\right) = 0 \text{ and } \sec\left(\frac{\pi}{2}\right) \text{ is undefined.}
\]
2. For \( \theta = \frac{3\pi}{2} \):
\[
\cos\left(\frac{3\pi}{2}\right) = 0 \text{ and } \sec\left(\frac{3\pi}{2}\right) \text{ is undefined.}
\]
Thus, the value of \( \theta \) on the interval \( [0, 2\pi) \) that satisfies the original equation is:
\[
\theta = \pi
\]
In conclusion, the solution to the equation \( \cos \theta = 1 + 2 \sec \theta \) in the interval \( [0, 2\pi) \) is:
\[
\boxed{\pi}
\]
\[
\sec \theta = \frac{1}{\cos \theta}
\]
Substituting this into the equation gives:
\[
\cos \theta = 1 + \frac{2}{\cos \theta}
\]
Next, we can multiply both sides of the equation by \( \cos \theta \) (noting that \( \cos \theta \neq 0 \), so we must check separately if \( \cos \theta = 0 \)):
\[
\cos^2 \theta = \cos \theta + 2
\]
Rearranging gives us a quadratic equation:
\[
\cos^2 \theta - \cos \theta - 2 = 0
\]
This can be factored as follows:
\[
(\cos \theta - 2)(\cos \theta + 1) = 0
\]
Setting each factor to zero gives:
1. \( \cos \theta - 2 = 0 \) which gives \( \cos \theta = 2 \) (not possible since cosine values are always between -1 and 1).
2. \( \cos \theta + 1 = 0 \) which gives \( \cos \theta = -1 \).
Next, we solve for \( \theta \) when \( \cos \theta = -1 \):
\[
\theta = \pi
\]
Now, we need to check if there are any angles where \( \cos \theta = 0 \) that would make our original assumption invalid. The values of \( \theta \) where \( \cos \theta = 0 \) are:
\[
\theta = \frac{\pi}{2}, \frac{3\pi}{2}
\]
However, substituting these values in the original equation:
1. For \( \theta = \frac{\pi}{2} \):
\[
\cos\left(\frac{\pi}{2}\right) = 0 \text{ and } \sec\left(\frac{\pi}{2}\right) \text{ is undefined.}
\]
2. For \( \theta = \frac{3\pi}{2} \):
\[
\cos\left(\frac{3\pi}{2}\right) = 0 \text{ and } \sec\left(\frac{3\pi}{2}\right) \text{ is undefined.}
\]
Thus, the value of \( \theta \) on the interval \( [0, 2\pi) \) that satisfies the original equation is:
\[
\theta = \pi
\]
In conclusion, the solution to the equation \( \cos \theta = 1 + 2 \sec \theta \) in the interval \( [0, 2\pi) \) is:
\[
\boxed{\pi}
\]
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