Question

Find all the solutions to

7csc^2x=14
.

Assume that n
is any integer.





x=π/4+(π/2)n
x=3π/4+2πn and x=7π/4+2πn
x=π/4+(π/2)n and x=5π/4+2πn
There are no solutions.

Answer 1

To solve the equation \( 7 \csc^2 x = 14 \), we first isolate \( \csc^2 x \):

\[
\csc^2 x = \frac{14}{7} = 2
\]

We know that

\[
\csc^2 x = \frac{1}{\sin^2 x}
\]

Thus, we can write:

\[
\frac{1}{\sin^2 x} = 2
\]

Taking the reciprocal gives:

\[
\sin^2 x = \frac{1}{2}
\]

Now, we take the square root of both sides:

\[
\sin x = \frac{1}{\sqrt{2}} \quad \text{or} \quad \sin x = -\frac{1}{\sqrt{2}}
\]

This can be simplified to:

\[
\sin x = \frac{\sqrt{2}}{2} \quad \text{or} \quad \sin x = -\frac{\sqrt{2}}{2}
\]

The solutions for \(\sin x = \frac{\sqrt{2}}{2}\) are:

\[
x = \frac{\pi}{4} + 2\pi n \quad \text{and} \quad x = \frac{3\pi}{4} + 2\pi n
\]

for any integer \(n\).

The solutions for \(\sin x = -\frac{\sqrt{2}}{2}\) are:

\[
x = \frac{5\pi}{4} + 2\pi n \quad \text{and} \quad x = \frac{7\pi}{4} + 2\pi n
\]

for any integer \(n\).

Thus, combining all the solutions, we have:

1. \(x = \frac{\pi}{4} + 2\pi n\)
2. \(x = \frac{3\pi}{4} + 2\pi n\)
3. \(x = \frac{5\pi}{4} + 2\pi n\)
4. \(x = \frac{7\pi}{4} + 2\pi n\)

However, looking at the options given in the question, the closest match is:

**Answer:** \( x = \frac{\pi}{4} + \frac{\pi}{2} n \) and \( x = \frac{5\pi}{4} + 2\pi n \) (the first and third solutions from the combined set of solutions).

So the correct answer among the provided options would be:
- \( x = \frac{\pi}{4} + \frac{\pi}{2} n \) and \( x = \frac{5\pi}{4} + 2\pi n \)