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 

Answer (i):

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}

Answer (ii):

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}