To solve the equation given:
\[ 2 \tan^2 x - \sec x = 1 \]
We'll use the identity:
\[ \sec^2 x = 1 + \tan^2 x \]
From the identity, we can express \(\tan^2 x\) in terms of \(\sec x\):
\[ \tan^2 x = \sec^2 x - 1 \]
Substituting this into the original equation:
\[ 2(\sec^2 x - 1) - \sec x = 1 \]
Expanding this:
\[ 2\sec^2 x - 2 - \sec x = 1 \]
Then we can rearrange the equation:
\[ 2\sec^2 x - \sec x - 3 = 0 \]
This is a quadratic equation in terms of \(\sec x\). Let \(y = \sec x\). The equation becomes:
\[ 2y^2 - y - 3 = 0 \]
To solve for \(y\) using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 2\), \(b = -1\), and \(c = -3\):
\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-3)}}{2(2)} \] \[ y = \frac{1 \pm \sqrt{1 + 24}}{4} \] \[ y = \frac{1 \pm \sqrt{25}}{4} \] \[ y = \frac{1 \pm 5}{4} \]
This gives us two potential solutions:
- \(y = \frac{6}{4} = \frac{3}{2}\)
- \(y = \frac{-4}{4} = -1\)
Thus, the possible values for \(\sec x\) are:
- \(\sec x = \frac{3}{2}\)
- \(\sec x = -1\)
Next, let's find the corresponding \(\tan x\):
Using \(\sec^2 x = 1 + \tan^2 x\):
-
For \(\sec x = \frac{3}{2}\):
\[ \sec^2 x = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \] \[ \frac{9}{4} = 1 + \tan^2 x \] \[ \tan^2 x = \frac{9}{4} - 1 = \frac{5}{4} \] \[ \tan x = \pm \frac{\sqrt{5}}{2} \]
-
For \(\sec x = -1\):
\[ \sec^2 x = (-1)^2 = 1 \] \[ 1 = 1 + \tan^2 x \] \[ \tan^2 x = 0 \] \[ \tan x = 0 \]
Now, we can analyze the given options:
- secant x equals negative 1 - True
- sec x = 3 - False (the solutions for sec are \(\frac{3}{2}\) and \(-1\))
- tan x = 3 - False (the possible values for \(\tan x\) are \(\frac{\sqrt{5}}{2}\) or \(0\))
- tangent x equals negative 1 - False (none of the values from our findings match \(-1\))
- secant x equals \(\frac{3}{2}\) - True
Thus, the correct answers are:
- secant x equals negative 1
- secant x equals \(\frac{3}{2}\)