If sin 3x = 1 and 0° ≤ 3x ≤ 90°, find the values of

(i) sin x

(ii) cos 2x

(iii) tan² x - sec² x.

Given,

⇒ sin 3x = 1

⇒ sin 3x = sin 90°

⇒ 3x = 90°

⇒ x = $\dfrac{90}{3}$

⇒ x = 30°.

(i) Substituting value of x in sin x, we get :

⇒ sin x = sin 30° = $\dfrac{1}{2}$.

Hence, sin x = $\dfrac{1}{2}$.

(ii) Substituting value of x in cos 2x, we get :

⇒ cos 2x = cos 60° = $\dfrac{1}{2}$.

Hence, cos x = $\dfrac{1}{2}$.

(iii) Substituting value of x in tan² x - sec² x, we get :

$\Rightarrow \text{tan}^2 x - \text{sec}^2 x = \text{tan}^2 60° - \text{sec}^2 60° \\[1em] = (\sqrt{3})^2 - (2)^2 \\[1em] = 3 - 4 \\[1em] = -1.$

Hence, tan² x - sec² x = -1.