Question
If 2 tangent squared x minus secant x equals 1, which of the following are true? Select all that apply.
(1 point)
Responses
secant x equals negative 1
Image with alt text: secant x equals negative 1
sec x = 3
sec x = 3
tan x = 3
tan x = 3
tangent x equals negative 1
Image with alt text: tangent x equals negative 1
secant x equals start fraction 3 over 2 end fraction
Image with alt text: secant x equals start fraction 3 over 2 end fraction
(1 point)
Responses
secant x equals negative 1
Image with alt text: secant x equals negative 1
sec x = 3
sec x = 3
tan x = 3
tan x = 3
tangent x equals negative 1
Image with alt text: tangent x equals negative 1
secant x equals start fraction 3 over 2 end fraction
Image with alt text: secant x equals start fraction 3 over 2 end fraction
Answers
GPT-4o mini
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 \( \tan x \):
1. Rewriting the equation:
\[
2 \tan^2 x - \sec x = 1
\]
Substituting \( \sec x = \sqrt{1 + \tan^2 x} \):
\[
2 \tan^2 x - \sqrt{1 + \tan^2 x} = 1
\]
2. Let's isolate \( \sqrt{1+\tan^2 x} \):
\[
\sqrt{1+\tan^2 x} = 2\tan^2 x - 1
\]
3. Squaring both sides to eliminate the square root:
\[
1 + \tan^2 x = (2\tan^2 x - 1)^2
\]
Expanding the right side gives:
\[
1 + \tan^2 x = 4\tan^4 x - 4\tan^2 x + 1
\]
4. Rearranging the equation:
\[
0 = 4\tan^4 x - 5\tan^2 x
\]
5. Factoring out \( \tan^2 x \):
\[
0 = \tan^2 x (4\tan^2 x - 5)
\]
This gives us two solutions:
1. \( \tan^2 x = 0 \Rightarrow \tan x = 0 \)
2. \( 4\tan^2 x - 5 = 0 \Rightarrow \tan^2 x = \frac{5}{4} \Rightarrow \tan x = \pm \frac{\sqrt{5}}{2} \)
Next, let’s evaluate the secant values using the identity \( \sec^2 x = 1 + \tan^2 x \):
- For \( \tan x = 0 \):
\[
\sec^2 x = 1 + 0 = 1 \implies \sec x = 1 \text{ or } -1
\]
- 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:
1. **secant x equals negative 1**: True (when \( \tan x = 0 \))
2. **sec x = 3**: False (not a solution)
3. **tan x = 3**: False (not a solution)
4. **tangent x equals negative 1**: False (not given as solution)
5. **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 \(\frac{3}{2}\)
Using the identity \( \sec^2 x = 1 + \tan^2 x \), we can express everything in terms of \( \tan x \):
1. Rewriting the equation:
\[
2 \tan^2 x - \sec x = 1
\]
Substituting \( \sec x = \sqrt{1 + \tan^2 x} \):
\[
2 \tan^2 x - \sqrt{1 + \tan^2 x} = 1
\]
2. Let's isolate \( \sqrt{1+\tan^2 x} \):
\[
\sqrt{1+\tan^2 x} = 2\tan^2 x - 1
\]
3. Squaring both sides to eliminate the square root:
\[
1 + \tan^2 x = (2\tan^2 x - 1)^2
\]
Expanding the right side gives:
\[
1 + \tan^2 x = 4\tan^4 x - 4\tan^2 x + 1
\]
4. Rearranging the equation:
\[
0 = 4\tan^4 x - 5\tan^2 x
\]
5. Factoring out \( \tan^2 x \):
\[
0 = \tan^2 x (4\tan^2 x - 5)
\]
This gives us two solutions:
1. \( \tan^2 x = 0 \Rightarrow \tan x = 0 \)
2. \( 4\tan^2 x - 5 = 0 \Rightarrow \tan^2 x = \frac{5}{4} \Rightarrow \tan x = \pm \frac{\sqrt{5}}{2} \)
Next, let’s evaluate the secant values using the identity \( \sec^2 x = 1 + \tan^2 x \):
- For \( \tan x = 0 \):
\[
\sec^2 x = 1 + 0 = 1 \implies \sec x = 1 \text{ or } -1
\]
- 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:
1. **secant x equals negative 1**: True (when \( \tan x = 0 \))
2. **sec x = 3**: False (not a solution)
3. **tan x = 3**: False (not a solution)
4. **tangent x equals negative 1**: False (not given as solution)
5. **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 \(\frac{3}{2}\)