I didn't understood q3b. Explain how to solve the equation given

First, we want to get $\cos \theta$ by itself on one side of the equation.

Divide both sides by 2:

The reference angle ($\alpha$) is the acute angle formed with the x-axis. To find it, we ignore the negative sign for a moment and consider $\cos \alpha = \frac{1}{2}$.

We know that $\cos(60^\circ) = \frac{1}{2}$, so:

The cosine function is negative in the second and third quadrants.

• In the first quadrant, all trigonometric functions are positive.
• In the second quadrant, only sine is positive (cosine is negative).
• In the third quadrant, only tangent is positive (cosine is negative).
• In the fourth quadrant, only cosine is positive.

We are looking for solutions for $\theta$ between $180^\circ$ and $360^\circ$. This range covers the third and fourth quadrants.
Since $\cos \theta$ must be negative (from Step 1), and our range includes the third and fourth quadrants, we must look for the angle in the third quadrant because cosine is negative there. (In the fourth quadrant, cosine is positive).

To find an angle in the third quadrant using the reference angle $\alpha$, we use the formula:

Substitute the reference angle $\alpha = 60^\circ$:

This angle $240^\circ$ is within the specified range of $180^\circ \le \theta \le 360^\circ$.

Therefore, the solution is: