Question

for 0° ≤ x < 360°, what are the solutions to cos(\frac{x}{2}) - sin(x) = 0?
{0°, 60°, 300°}
{0°,120°, 240°}
{60°, 180°, 300°}
{120°,180°, 240°}

Explanation:

Step1: Use double - angle formula

Recall $\sin(x)=2\sin(\frac{x}{2})\cos(\frac{x}{2})$. The equation $\cos(\frac{x}{2})-\sin(x) = 0$ becomes $\cos(\frac{x}{2})-2\sin(\frac{x}{2})\cos(\frac{x}{2})=0$.

Step2: Factor out $\cos(\frac{x}{2})$

We get $\cos(\frac{x}{2})(1 - 2\sin(\frac{x}{2}))=0$.

Step3: Set each factor equal to zero

Case 1: $\cos(\frac{x}{2})=0$. Then $\frac{x}{2}=90^{\circ}+180^{\circ}n$, $x = 180^{\circ}+360^{\circ}n$. For $0^{\circ}\leq x<360^{\circ}$, when $n = 0$, $x = 180^{\circ}$.
Case 2: $1 - 2\sin(\frac{x}{2})=0$, so $\sin(\frac{x}{2})=\frac{1}{2}$. Then $\frac{x}{2}=30^{\circ}+360^{\circ}n$ or $\frac{x}{2}=150^{\circ}+360^{\circ}n$.
For $\frac{x}{2}=30^{\circ}+360^{\circ}n$, $x = 60^{\circ}+720^{\circ}n$. For $n = 0$, $x = 60^{\circ}$.
For $\frac{x}{2}=150^{\circ}+360^{\circ}n$, $x = 300^{\circ}+720^{\circ}n$. For $n = 0$, $x = 300^{\circ}$.
So the solutions in the range $0^{\circ}\leq x<360^{\circ}$ are $x = 60^{\circ},180^{\circ},300^{\circ}$.

Answer:

C. $\{60^{\circ}, 180^{\circ}, 300^{\circ}\}$