Question
Find all the solutions to
2+7csc^2x=12+2csc^2x
.
Assume that n
is any integer.
x=3π/4+2πn and x=7π/4+2πn
x=π/4+(π/2)n and x=5π/4+2πn
x=π/4+(π/2)n
There are no solutions.
2+7csc^2x=12+2csc^2x
.
Assume that n
is any integer.
x=3π/4+2πn and x=7π/4+2πn
x=π/4+(π/2)n and x=5π/4+2πn
x=π/4+(π/2)n
There are no solutions.
Answers
GPT-4o mini
To solve the equation
\[
2 + 7\csc^2 x = 12 + 2\csc^2 x,
\]
we first rearrange the equation by bringing all terms involving \(\csc^2 x\) on one side:
\[
7\csc^2 x - 2\csc^2 x = 12 - 2.
\]
This simplifies to:
\[
5\csc^2 x = 10.
\]
Next, we divide both sides by 5:
\[
\csc^2 x = 2.
\]
Since \(\csc^2 x = \frac{1}{\sin^2 x}\), we can rewrite this as:
\[
\frac{1}{\sin^2 x} = 2,
\]
which implies that:
\[
\sin^2 x = \frac{1}{2}.
\]
Taking the square root of both sides gives:
\[
\sin x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}.
\]
The solutions for \(\sin x = \frac{\sqrt{2}}{2}\) occur at:
\[
x = \frac{\pi}{4} + 2\pi n
\]
and
\[
x = \frac{3\pi}{4} + 2\pi n.
\]
The solutions for \(\sin x = -\frac{\sqrt{2}}{2}\) occur at:
\[
x = \frac{5\pi}{4} + 2\pi n
\]
and
\[
x = \frac{7\pi}{4} + 2\pi n.
\]
Combining all of this, the complete set of solutions is:
\[
x = \frac{\pi}{4} + 2\pi n, \quad x = \frac{3\pi}{4} + 2\pi n, \quad x = \frac{5\pi}{4} + 2\pi n, \quad x = \frac{7\pi}{4} + 2\pi n.
\]
Ensuring this matches any of the given options, we can simplify and group them into specific forms. The simplest representation of the solutions is:
\[
x = \frac{\pi}{4} + \frac{\pi}{2} n \quad (n \text{ being even or odd for the other solutions})
\text{ and }
\]
\[
x = \frac{3\pi}{4} + 2\pi n.
\]
The significant options that show combinations of the possible solutions indicate the presence of:
1. \( x = \frac{\pi}{4} + \frac{\pi}{2} n \text{ includes }\)
2. \( x = \frac{5\pi}{4} + 2\pi n \) and similarly for others.
Thus, grouping these solutions, we conclude:
\(\left( x = \frac{\pi}{4} + \frac{\pi}{2} n \text{ and } x = \frac{5\pi}{4} + 2\pi n \text{ are inclusive of the main solutions }\right)\).
Finally, noticing patterns in terms of \( \pi/4 + \frac{\pi}{2} n \):
The option that fits the solutions pattern is:
\[
\boxed{x = \frac{\pi}{4} + \frac{\pi}{2} n}
\]
\[
2 + 7\csc^2 x = 12 + 2\csc^2 x,
\]
we first rearrange the equation by bringing all terms involving \(\csc^2 x\) on one side:
\[
7\csc^2 x - 2\csc^2 x = 12 - 2.
\]
This simplifies to:
\[
5\csc^2 x = 10.
\]
Next, we divide both sides by 5:
\[
\csc^2 x = 2.
\]
Since \(\csc^2 x = \frac{1}{\sin^2 x}\), we can rewrite this as:
\[
\frac{1}{\sin^2 x} = 2,
\]
which implies that:
\[
\sin^2 x = \frac{1}{2}.
\]
Taking the square root of both sides gives:
\[
\sin x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}.
\]
The solutions for \(\sin x = \frac{\sqrt{2}}{2}\) occur at:
\[
x = \frac{\pi}{4} + 2\pi n
\]
and
\[
x = \frac{3\pi}{4} + 2\pi n.
\]
The solutions for \(\sin x = -\frac{\sqrt{2}}{2}\) occur at:
\[
x = \frac{5\pi}{4} + 2\pi n
\]
and
\[
x = \frac{7\pi}{4} + 2\pi n.
\]
Combining all of this, the complete set of solutions is:
\[
x = \frac{\pi}{4} + 2\pi n, \quad x = \frac{3\pi}{4} + 2\pi n, \quad x = \frac{5\pi}{4} + 2\pi n, \quad x = \frac{7\pi}{4} + 2\pi n.
\]
Ensuring this matches any of the given options, we can simplify and group them into specific forms. The simplest representation of the solutions is:
\[
x = \frac{\pi}{4} + \frac{\pi}{2} n \quad (n \text{ being even or odd for the other solutions})
\text{ and }
\]
\[
x = \frac{3\pi}{4} + 2\pi n.
\]
The significant options that show combinations of the possible solutions indicate the presence of:
1. \( x = \frac{\pi}{4} + \frac{\pi}{2} n \text{ includes }\)
2. \( x = \frac{5\pi}{4} + 2\pi n \) and similarly for others.
Thus, grouping these solutions, we conclude:
\(\left( x = \frac{\pi}{4} + \frac{\pi}{2} n \text{ and } x = \frac{5\pi}{4} + 2\pi n \text{ are inclusive of the main solutions }\right)\).
Finally, noticing patterns in terms of \( \pi/4 + \frac{\pi}{2} n \):
The option that fits the solutions pattern is:
\[
\boxed{x = \frac{\pi}{4} + \frac{\pi}{2} n}
\]