Question

Find the value of x if cos2x = sin 60° cos 30° - cos 60° sin 30°.

Answer 1

Answer:

see attached file for your answer

Answer 2

We know that,

$\dagger\quad \bf\sin 60^{\circ} = \dfrac{\sqrt{3}}{2} = \cos 30^{\circ}$

$\dagger\quad \bf\sin 30^{\circ} = \dfrac{1}{2} = \cos 60^{\circ}$

Hence, substituting the values we get,

$\dashrightarrow\rm \cos 2x = \sin 60^{\circ} \cos 30^{\circ} - \cos 60^{\circ} \sin 30^{\circ}$

$\dashrightarrow\rm \cos 2x = \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} - \dfrac{1}{2}\times \dfrac{1}{2}$

$\dashrightarrow\rm \cos 2x = \dfrac{(\sqrt{3})^2}{(2)^2}- \dfrac{1}{4}$

$\dashrightarrow\rm \cos 2x = \dfrac{3}{4}- \dfrac{1}{4}$

$\dashrightarrow \rm\cos 2x = \dfrac{2}{4}$

$\dashrightarrow\rm \cos 2x = \dfrac{1}{2}$

Since, cos 60° = 1/2, hence,

$\dashrightarrow \rm\cos 2x = \dfrac{1}{2} = \cos 60^{\circ}$

$\dashrightarrow \rm\cos 2x = \cos 60^{\circ}$

$\dashrightarrow\rm 2x = 60^{\circ}$

$\dashrightarrow \rm x = \dfrac{60^{\circ}}{2}$

$\dashrightarrow \boxed{ \bf x =30^{\circ}}$