Question Number 38881 by Rio Mike last updated on 30/Jun/18
$${solve}\:{for}\:\mathrm{0}\:\leqslant\:\theta\:\leqslant\pi,\:{the}\:{equations} \\ $$$$\left.{a}\right)_{} \:\mathrm{cos}\:\left(\mathrm{2}\theta\:−\:\frac{\pi}{\mathrm{2}}\right)=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left.{b}\right)\:\mathrm{cos}\:\theta\:−\:\sqrt{\mathrm{3}}\:{sin}\:\theta\:=\:\mathrm{0} \\ $$
Answered by MrW3 last updated on 30/Jun/18
$$\mathrm{0}\leqslant\theta\leqslant\pi \\ $$$$−\frac{\pi}{\mathrm{2}}\leqslant\mathrm{2}\theta−\frac{\pi}{\mathrm{2}}\leqslant\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$$\left({a}\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{2}\theta\:−\:\frac{\pi}{\mathrm{2}}\right)=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}\theta\:−\:\frac{\pi}{\mathrm{2}}=\:−\frac{\pi}{\mathrm{3}},\:\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow\theta=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}\right)=\frac{\pi}{\mathrm{12}} \\ $$$$\Rightarrow\theta=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)=\frac{\mathrm{5}\pi}{\mathrm{12}} \\ $$$$\left({b}\right) \\ $$$$\mathrm{cos}\:\theta\:−\:\sqrt{\mathrm{3}}\:{sin}\:\theta\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\theta=\frac{\pi}{\mathrm{6}} \\ $$