Solution method:

1. Find the reference angle: $\theta_r = \arcsin(|a|)$
2. Determine which quadrants based on the sign of $a$
3. Find all solutions in $[0, 2\pi)$
4. Add $2\pi n$ for general solution

General solution patterns:

• If $\sin(x) = a$: $x = \arcsin(a) + 2\pi n$ or $x = \pi - \arcsin(a) + 2\pi n$
• If $\cos(x) = a$: $x = \arccos(a) + 2\pi n$ or $x = -\arccos(a) + 2\pi n$
• If $\tan(x) = a$: $x = \arctan(a) + \pi n$

Type 2: Quadratic Equations

Example form: $\sin^2(x) + b\sin(x) + c = 0$

Solution method:

1. Substitute $u = \sin(x)$ (or the relevant trig function)
2. Solve the quadratic equation for $u$
3. Solve $\sin(x) = u$ for each value of $u$
4. Check that solutions are in the domain

Type 3: Equations with Multiple Angles

Example form: $\sin(2x) = \frac{1}{2}$

Solution method:

1. Let $u = 2x$ (or the multiple angle)
2. Solve $\sin(u) = \frac{1}{2}$
3. Divide by the coefficient to find $x$
4. Consider the extended period

Type 4: Equations with Multiple Functions

Example form: $\sin(x) + \cos(x) = 1$

Solution methods:

• Use Pythagorean identities to express in terms of one function
• Square both sides (check for extraneous solutions)
• Use sum-to-product or product-to-sum identities

Key Identities for Solving

Pythagorean Identities

• $\sin^2(x) + \cos^2(x) = 1$
• $1 + \tan^2(x) = \sec^2(x)$
• $1 + \cot^2(x) = \csc^2(x)$

Double Angle Formulas

• $\sin(2x) = 2\sin(x)\cos(x)$
• $\cos(2x) = \cos^2(x) - \sin^2(x) = 2\cos^2(x) - 1 = 1 - 2\sin^2(x)$
• $\tan(2x) = \frac{2\tan(x)}{1 - \tan^2(x)}$

Half Angle Formulas

• $\sin^2(x) = \frac{1 - \cos(2x)}{2}$
• $\cos^2(x) = \frac{1 + \cos(2x)}{2}$

Interval Considerations

• Standard interval: $[0, 2\pi)$ or $[0°, 360°)$
• Extended intervals: May need to consider multiple periods
• General solution: Include $+ 2\pi n$ (or $+ \pi n$ for tangent) where $n$ is an integer

Checking Solutions

Always verify solutions by:

1. Substituting back into the original equation
2. Checking domain restrictions (e.g., no division by zero)
3. Eliminating extraneous solutions from squaring

Common Reference Angles

This occurs at $x = \frac{\pi}{3}$ (Quadrant I) and $x = \frac{2\pi}{3}$ (Quadrant II)

Solutions: $x = \frac{\pi}{3}, \frac{2\pi}{3}$

Part (b): $\cos^2 x = \frac{1}{4}$

$\cos x = \pm\frac{1}{2}$

For $\cos x = \frac{1}{2}$: $x = \frac{\pi}{3}, \frac{5\pi}{3}$

For $\cos x = -\frac{1}{2}$: $x = \frac{2\pi}{3}, \frac{4\pi}{3}$

Solutions: $x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$

Part (c): $2\cos x - 1 = 0$

$\cos x = \frac{1}{2}$

Solutions: $x = \frac{\pi}{3}, \frac{5\pi}{3}$

Step 1: Factor or use quadratic formula

Let $u = \cos(x)$: $2u^2 - u - 1 = 0$

Factor: $(2u + 1)(u - 1) = 0$

Step 2: Solve for $u$ $2u + 1 = 0 \quad \text{or} \quad u - 1 = 0$ $u = -\frac{1}{2} \quad \text{or} \quad u = 1$

Step 3: Solve for $x$

Case 1: $\cos(x) = -\frac{1}{2}$

Cosine is negative in Quadrants II and III. Reference angle: $\arccos(\frac{1}{2}) = \frac{\pi}{3}$

• Quadrant II: $x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
• Quadrant III: $x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$

Case 2: $\cos(x) = 1$

$x = 0$ (or $2\pi$, but we use $[0, 2\pi)$)

Answer: $x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}$

Verification:

• At $x = 0$: $2(1)^2 - 1 - 1 = 0$ ✓
• At $x = \frac{2\pi}{3}$: $2(-\frac{1}{2})^2 - (-\frac{1}{2}) - 1 = \frac{1}{2} + \frac{1}{2} - 1 = 0$ ✓
• At $x = \frac{4\pi}{3}$: $2(-\frac{1}{2})^2 - (-\frac{1}{2}) - 1 = 0$ ✓

$2\sin x \cos x - \cos x = 0$

$\cos x(2\sin x - 1) = 0$

This gives us two cases:

Case 1: $\cos x = 0$

In $[0, 2\pi)$: $x = \frac{\pi}{2}, \frac{3\pi}{2}$

Case 2: $2\sin x - 1 = 0$

$\sin x = \frac{1}{2}$

In $[0, 2\pi)$: $x = \frac{\pi}{6}, \frac{5\pi}{6}$

All solutions: $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$

Step 2: Move all terms to one side $2\sin(x)\cos(x) - \sin(x) = 0$

Step 3: Factor out $\sin(x)$ $\sin(x)(2\cos(x) - 1) = 0$

Step 4: Solve each factor

Factor 1: $\sin(x) = 0$

In $[0, 2\pi)$: $x = 0, \pi$

Factor 2: $2\cos(x) - 1 = 0$ $\cos(x) = \frac{1}{2}$

Cosine is positive in Quadrants I and IV. Reference angle: $\arccos(\frac{1}{2}) = \frac{\pi}{3}$

• Quadrant I: $x = \frac{\pi}{3}$
• Quadrant IV: $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$

Answer: $x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}$

Verification:

• At $x = 0$: $\sin(0) = \sin(0) = 0$ ✓
• At $x = \frac{\pi}{3}$: $\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ ✓
• At $x = \pi$: $\sin(2\pi) = \sin(\pi) = 0$ ✓
• At $x = \frac{5\pi}{3}$: $\sin(\frac{10\pi}{3}) = \sin(\frac{4\pi}{3}) = -\frac{\sqrt{3}}{2}$ and ✓

Note: We reduced $\sin(\frac{10\pi}{3})$ by subtracting $2\pi$: $\frac{10\pi}{3} - 2\pi = \frac{4\pi}{3}$