If 4 cos² x = 3 and x is an acute angle; find the value of :

(i) x

(ii) cos² x + cot² x

(iii) cos 3x

(iv) sin 2x

(i) 4 cos² x = 3

⇒ cos² x = $\dfrac{3}{4}$

⇒ cos x = $\sqrt\dfrac{3}{4}$

⇒ cos x = $\dfrac{\sqrt3}{2}$

⇒ cos x = cos 30°

Hence, x = 30°.

(ii) cos x = cos 30° = $\dfrac{\sqrt3}{2}$

cot x = cot 30° = $\sqrt3$

Now, cos² x + cot² x

$= \Big(\dfrac{\sqrt3}{2}\Big)^2 + (\sqrt3)^2\\[1em] = \dfrac{3}{4} + 3\\[1em] = \dfrac{3}{4} + \dfrac{12}{4}\\[1em] = \dfrac{3 + 12}{4}\\[1em] = \dfrac{15}{4}\\[1em] = 3\dfrac{3}{4}$

Hence, cos² x + cot² x = $3\dfrac{3}{4}$.

(iii) cos 3x = cos (3 x 30°)

= cos 90° = 0

Hence, cos 3x = 0.

(iv) sin 2x = sin (2 x 30°)

= sin 60° = $\dfrac{\sqrt3}{2}$

Hence, sin 2x = $\dfrac{\sqrt3}{2}$.