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Let's solve the three parts of the question one by one.

### Given range for \( x \): \( -\pi < x \leq \pi \)

### (i) Solve \( \cos(2x) = \frac{\sqrt{3}}{2} \)
The cosine equation is given by:
\[
\cos(2x) = \frac{\sqrt{3}}{2}
\]

We know that \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \) corresponds to angles where cosine has this value. From the unit circle, this occurs at:
\[
2x = \pm \frac{\pi}{6} + 2k\pi \quad \text{for some integer } k
\]
However, since we are interested in \( -\pi < x \leq \pi \), we need to solve for \( x \).

Dividing both sides by 2, we get:
\[
x = \pm \frac{\pi}{12} + k\pi
\]

Now, let's find the solutions in the interval \( -\pi < x \leq \pi \):

- For \( 2x = \frac{\pi}{6} \), we get \( x = \frac{\pi}{12} \)
- For \( 2x = -\frac{\pi}{6} \), we get \( x = -\frac{\pi}{12} \)

These are the only solutions in the interval.

### (ii) Solve \( \tan^2(x) = \frac{1}{3} \)
Take the square root on both sides:
\[
\tan(x) = \pm \frac{1}{\sqrt{3}}
\]

From the unit circle, we know that:
\[
\tan(x) = \frac{1}{\sqrt{3}} \quad \text{at} \quad x = \pm \frac{\pi}{6} + k\pi
\]

Now, let's find the solutions for \( -\pi < x \leq \pi \):
- For \( x = \frac{\pi}{6} \)
- For \( x = -\frac{\pi}{6} \)
- For \( x = \frac{\pi}{6} + \pi = \frac{7\pi}{6} \) (which is not within the required range)
- For \( x = -\frac{\pi}{6} + \pi = \frac{5\pi}{6} \)

Thus, the solutions are \( x = \frac{\pi}{6}, -\frac{\pi}{6}, \frac{5\pi}{6} \).

### (iii) Solve \( 2 \sin^2(x) = \sin(x) + 1 \)
Rearrange the equation:
\[
2\sin^2(x) - \sin(x) - 1 = 0
\]

This is a quadratic equation in \( \sin(x) \). Let \( y = \sin(x) \), then the equation becomes:
\[
2y^2 - y - 1 = 0
\]

Solve this quadratic using the quadratic formula:
\[
y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}
\]

Thus:
\[
y = \frac{1 + 3}{4} = 1 \quad \text{or} \quad y = \frac{1 - 3}{4} = -\frac{1}{2}
\]

So, \( \sin(x) = 1 \) or \( \sin(x) = -\frac{1}{2} \).

Now, find the solutions for \( x \) in \( -\pi < x \leq \pi \):
- \( \sin(x) = 1 \) at \( x = \frac{\pi}{2} \)
- \( \sin(x) = -\frac{1}{2} \) at \( x = -\frac{\pi}{6}, \frac{7\pi}{6} \) (but we reject \( \frac{7\pi}{6} \) since it is outside the range)

Thus, the solutions are \( x = \frac{\pi}{2}, -\frac{\pi}{6} \).

---

Do you want any further details or have questions?

Here are five related questions:
1. How do you solve quadratic trigonometric equations like in part (iii)?
2. What are the key points to remember for solving \( \cos(2x) = \frac{\sqrt{3}}{2} \)?
3. What is the importance of the unit circle in solving trigonometric equations?
4. How can the general solution for tangent functions be applied in various intervals?
5. Why is it important to check the interval when solving trigonometric equations?

**Tip:** When solving trigonometric equations, always refer to the unit circle and consider the given interval carefully.

Solve for x, where -π < x ≤ π, giving your answer exactly where appropriate:
(i) cos(2x) = √3/2
(ii) tan^2(x) = 1/3
(iii) 2sin^2(x) = sin(x) + 1

Learn how to solve trigonometric equations involving cosine and tangent functions within the interval -π < x ≤ π. This step-by-step guide also covers solving a quadratic trigonometric equation. Ideal for students in Grades 10-12.

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