$\sin(2x-60 ^{\circ}) = 0.5$ for $x\in [0^\circ, 360^\circ]$

I have worked out the answers $45^\circ$, $135^\circ$, $225^\circ$, $315^\circ$; although some of these may be wrong. Where have I gone wrong (if I have)? And are there any other answers (between $0^\circ$ and $360^\circ$ inclusive) that I have missed out?

2 Answers

$\sin30^\circ = \sin150^\circ = \sin(30^\circ+360^\circ) = \sin(150^\circ+360^\circ) = \dfrac{1}{2}$ so you need to seek $x$ values where

$$2x - 60 = 30 \implies x = 45^\circ$$

$$2x-60 = 150 \implies x = 105^\circ$$

But notice that since $x \in [0^\circ,360^\circ]$, we also may have

$$2x-60 = 390 \implies x = 225^\circ$$

$$2x-60 = 510 \implies x = 285^\circ$$

Note that $$\sin y=\frac12\iff y=30°+2k\cdot180°,150°+2k\cdot180°\implies 2x=90°+k360°,210°+k360°$$

thus since $x\in[0,360°]$

$$x=45°,225°$$

$$x=105°,285°$$

• sorry for the initial confusion, now it should be ok!
– user
Feb 8 '18 at 21:32