To solve the equation \( 2 \tan^2 x - \sec x = 1 \), let's first rewrite it in terms of only one trigonometric function.
Using the identity \( \sec^2 x = 1 + \tan^2 x \), we can express everything in terms of \( \tan x \):
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Rewriting the equation: \[ 2 \tan^2 x - \sec x = 1 \] Substituting \( \sec x = \sqrt{1 + \tan^2 x} \): \[ 2 \tan^2 x - \sqrt{1 + \tan^2 x} = 1 \]
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Let's isolate \( \sqrt{1+\tan^2 x} \): \[ \sqrt{1+\tan^2 x} = 2\tan^2 x - 1 \]
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Squaring both sides to eliminate the square root: \[ 1 + \tan^2 x = (2\tan^2 x - 1)^2 \] Expanding the right side gives: \[ 1 + \tan^2 x = 4\tan^4 x - 4\tan^2 x + 1 \]
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Rearranging the equation: \[ 0 = 4\tan^4 x - 5\tan^2 x \]
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Factoring out \( \tan^2 x \): \[ 0 = \tan^2 x (4\tan^2 x - 5) \]
This gives us two solutions:
- \( \tan^2 x = 0 \Rightarrow \tan x = 0 \)
- \( 4\tan^2 x - 5 = 0 \Rightarrow \tan^2 x = \frac{5}{4} \Rightarrow \tan x = \pm \frac{\sqrt{5}}{2} \)
Next, let’s evaluate the secant values using the identity \( \sec^2 x = 1 + \tan^2 x \):
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For \( \tan x = 0 \): \[ \sec^2 x = 1 + 0 = 1 \implies \sec x = 1 \text{ or } -1 \]
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For \( \tan x = \frac{\sqrt{5}}{2} \): \[ \sec^2 x = 1 + \frac{5}{4} = \frac{9}{4} \implies \sec x = \frac{3}{2} \text{ or } -\frac{3}{2} \]
Now, let's see which values from the options provided can be true:
- secant x equals negative 1: True (when \( \tan x = 0 \))
- sec x = 3: False (not a solution)
- tan x = 3: False (not a solution)
- tangent x equals negative 1: False (not given as solution)
- secant x equals \(\frac{3}{2}\): True (when \( \tan x = \frac{\sqrt{5}}{2} \))
The true statements are:
- secant x equals negative 1
- secant x equals \(\frac{3}{2}\)