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We are tasked with solving three trigonometric equations, given specific conditions on the angles. Here's the breakdown:

1. **Conditions:**
   - The variable \( n \) is constrained to the interval \( 0^\circ \leq n \leq 90^\circ \).
   - The variable \( x \) is constrained to the interval \( 90^\circ \leq x \leq 360^\circ \).
   - We need to solve each equation in terms of \( n \) and find solutions for \( x \).

The equations are:
a) \( \sin x = \sin n \)  
b) \( \cos x = \cos n \)  
c) \( \tan x = \tan n \)

### Solutions:

#### a) \( \sin x = \sin n \)

For the sine function, \( \sin \theta \) is positive in the first and second quadrants, but since \( x \) is restricted to \( 90^\circ \leq x \leq 360^\circ \), we only consider the second quadrant and fourth quadrant.

- In the second quadrant, the general solution for \( \sin x = \sin n \) is:
  \[
  x = 180^\circ - n
  \]
  
- In the fourth quadrant, the general solution is:
  \[
  x = 360^\circ + (-n) = 360^\circ - n
  \]

Thus, for \( \sin x = \sin n \), the solutions are:
\[
x = 180^\circ - n \quad \text{or} \quad x = 360^\circ - n
\]

#### b) \( \cos x = \cos n \)

For the cosine function, \( \cos \theta \) is positive in the first and fourth quadrants, and negative in the second and third quadrants. Since \( x \) is restricted to \( 90^\circ \leq x \leq 360^\circ \), we need to consider the second and third quadrants for \( \cos x = \cos n \).

- In the second quadrant, the general solution for \( \cos x = \cos n \) is:
  \[
  x = 360^\circ - n
  \]

- In the third quadrant, the general solution is:
  \[
  x = 180^\circ + n
  \]

Thus, for \( \cos x = \cos n \), the solutions are:
\[
x = 360^\circ - n \quad \text{or} \quad x = 180^\circ + n
\]

#### c) \( \tan x = \tan n \)

For the tangent function, \( \tan \theta \) is periodic with a period of \( 180^\circ \). This means that the solutions for \( \tan x = \tan n \) can occur at intervals of \( 180^\circ \).

- The general solution for \( \tan x = \tan n \) is:
  \[
  x = 180^\circ k + n
  \]
  where \( k \) is an integer.

Since \( x \) is restricted to \( 90^\circ \leq x \leq 360^\circ \), we can choose values of \( k \) to fit this range:
- For \( k = 1 \), we get:
  \[
  x = 180^\circ \times 1 + n = 180^\circ + n
  \]
- For \( k = 2 \), we get:
  \[
  x = 180^\circ \times 2 + n = 360^\circ + n \quad \text{(but this exceeds 360°, so we discard this)}
  \]

Thus, for \( \tan x = \tan n \), the solution is:
\[
x = 180^\circ + n
\]

### Summary of Solutions:
a) \( x = 180^\circ - n \) or \( x = 360^\circ - n \)  
b) \( x = 360^\circ - n \) or \( x = 180^\circ + n \)  
c) \( x = 180^\circ + n \)

Would you like more details on any part of this, or do you have any questions?

Here are some related questions for further practice:

1. How would the solutions change if the interval for \( x \) was \( 0^\circ \leq x \leq 360^\circ \)?
2. What happens to the solutions if \( n \) is restricted to \( 0^\circ \leq n \leq 180^\circ \)?
3. How would you solve \( \sin x = \cos n \) for \( x \)?
4. What are the general solutions for \( \sin x = 1/2 \) for any \( x \)?
5. Can you express \( \cos(180^\circ - n) \) in terms of \( \cos n \)?

**Tip:** Always remember that sine, cosine, and tangent functions repeat their values periodically, which can help you find solutions over different intervals.

Solve the following equations when 90° ≤ x ≤ 360°, giving your answers in terms of angle n:
a) sin x = sin n
b) cos x = cos n
c) tan x = tan n

This page provides step-by-step solutions for solving trigonometric equations involving sine, cosine, and tangent functions in the angle range 90° ≤ x ≤ 360°, with results given in terms of angle n. Suitable for high school students.

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