Solve, for $0 \leq x < 360^{\circ},$ (a) $\sin(x - 20^{\circ}) = -\frac{1}{\sqrt{2}}$ (b) $\cos 3x = -\frac{1}{2}$ - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 2

To solve this equation, we first recognize that the cosine function is negative in the second and third quadrants. The reference angle for $-\frac{1}{2}$ is $120^{\circ}$, leading us to the following equations:

1. $3x = 120^{\circ} + 360^{\circ}k$ for $k \in \mathbb{Z}$
2. $3x = 240^{\circ} + 360^{\circ}k$ for $k \in \mathbb{Z}$

Dividing by 3 gives:

1. $x = 40^{\circ} + 120^{\circ}k$
2. $x = 80^{\circ} + 120^{\circ}k$

Next, substituting integer values for $k$ such that $0 \leq x < 360^{\circ}$:

For $k = 0$:

• From $x = 40^{\circ}$, we have $x = 40^{\circ}$
• From $x = 80^{\circ}$, we have $x = 80^{\circ}$

For $k = 1$:

• From $x = 40^{\circ}$, we have $x = 160^{\circ}$
• From $x = 80^{\circ}$, we have $x = 200^{\circ}$

For $k = 2$:

• From $x = 40^{\circ}$, we have $x = 280^{\circ}$
• From $x = 80^{\circ}$, we have $x = 320^{\circ}$

Thus, the solutions for part (b) are:

• $x = 40^{\circ}$
• $x = 80^{\circ}$
• $x = 160^{\circ}$
• $x = 200^{\circ}$
• $x = 280^{\circ}$
• $x = 320^{\circ}$