Solve each of the following equations for $0^{0}\leq x\leq 180^{0}$(i) $\sin 2x=0.5$

(ii) $2\sin ^{2}x=2-\sqrt {3}\cos x$

Solve $\sin 2x=0.5$

Assuming u=2x,

$\sin u=0.5 \\ \\ u=\sin ^{-1}0.5 \\ \\ u=30^{0}$

Now $\sin \left( 180^{0}-u\right) =\sin u$

$\therefore u=180^{0} - 30^{0}=150^{0} \\ \\ \therefore u= 30^{0}, 150^{0}$

As $u=2x$

$\therefore 2x= 30^{0},150^{0}$

or, $x=15^{0}, 75^{0}$

Solve $2\sin ^{2}x=2-\sqrt {3}\cos x$

We know that, $\sin ^{2}x+\cos ^{2}x=1 \\ \therefore \sin ^{2}x=1-\cos ^{2}x$

Substituting the value of $\sin ^{2}x$ in $2\sin ^{2}x=2-\sqrt {3}\cos x$, we get,

$2\times \left( 1-\cos ^{2}x\right) = 2-\sqrt {3}\cos x$

$\therefore 2-2\cos ^{2}x = 2-\sqrt {3}\cos x$

or, $2-2\cos ^{2}x - 2 + \sqrt {3}\cos x = 0$

or, $2\cos ^{2}x - \sqrt {3}\cos x = 0$

or, $\cos x\left( 2\cos x-\sqrt {3}\right) =0$

or, $\cos x\left( 2\cos x-\sqrt {3}\right) =0$

or, $2\cos x-\sqrt {3}=0$ or $\cos x=0$

or, $\cos x=\dfrac {\sqrt {3}}{2}$ or $\cos x=0$

$\therefore x=\cos ^{-1}\left( \dfrac {\sqrt {3}}{2}\right)$ or, $x=\cos ^{-1}0$

$\therefore x=30^{0}, x=90^{0}$