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:
- \( \cos \theta - 2 = 0 \) which gives \( \cos \theta = 2 \) (not possible since cosine values are always between -1 and 1).
- \( \cos \theta + 1 = 0 \) 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:
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For \( \theta = \frac{\pi}{2} \): \[ \cos\left(\frac{\pi}{2}\right) = 0 \text{ and } \sec\left(\frac{\pi}{2}\right) \text{ is undefined.} \]
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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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