Solve \(sin(2x+40^{\circ}) = 0.6 \) in the interval \( \quad 0^{\circ} \leqslant x \leqslant 180^{\circ}\)
(solutions to 2 d.p.)
Our modified interval is \( (2 \times 0^{\circ})+40^{\circ} \leqslant 2x+40^{\circ} \leqslant (2 \times 180^{\circ})+40^{\circ} \)
\( 40^{\circ} \leqslant 2x+40^{\circ} \leqslant 400^{\circ} \)
\begin{align} \\ & and\ we\ know \\ \\ sin^{-1}(0.6) & = 2x+40^{\circ} \\ \\ So \quad 2x+40^{\circ} & = 36.869^{\circ}, \, 143.131^{\circ}, \, 396.131^{\circ} \\ \\ & But\ 36.869^{\circ}\ is\ outside\ our\ interval\ so \\ \\ 2x+ 40^{\circ} & = 143.131^{\circ} \\ x & =51.57^{\circ} ( to\ 2\ d.p.) \\ & and \\ 2x+40^{\circ}& =396.131^{\circ} \\ x & =178.07^{\circ} (to\ 2\ d.p) \end{align}