If 4 cos² x° - 1 = 0 and 0 ≤ x ≤ 90, find

(i) x

(ii) sin² x° + cos² x°

(iii) cos² x° - sin² x°.

(i) Given,

$\Rightarrow \text{4 cos}^2 x° - 1 = 0 \\[1em] \Rightarrow \text{4 cos}^2 x° = 1 \\[1em] \Rightarrow \text{cos}^2 x° = \dfrac{1}{4} \\[1em] \Rightarrow \text{cos x°} = \sqrt{\dfrac{1}{4}} \\[1em] \Rightarrow \text{cos x°} = \pm \dfrac{1}{2}$

Since, x is an acute angle.

$\therefore \text{cos x°} = \dfrac{1}{2}$.

⇒ cos x° = cos 60°

⇒ x = 60.

Hence, x = 60.

(ii) Substituting value of x in sin² x° + cos² x° we get :

$\text{sin}^2 x° + \text{cos}^2 x° = \text{sin}^2 60° + \text{cos}^2 60° \\[1em] = \Big(\dfrac{\sqrt{3}}{2}\Big)^2 + \Big(\dfrac{1}{2}\Big)^2 \\[1em] = \dfrac{3}{4} + \dfrac{1}{4} \\[1em] = \dfrac{4}{4} \\[1em] = 1.$

Hence, sin² x° + cos² x° = 1.

(iii) Substituting value of x in cos² x° - sin² x° we get :

$\text{cos}^2 x° - \text{sin}^2 x° = \text{cos}^2 60° - \text{sin}^2 60° \\[1em] = \Big(\dfrac{1}{2}\Big)^2 - \Big(\dfrac{\sqrt{3}}{2}\Big)^2 \\[1em] = \dfrac{1}{4} - \dfrac{3}{4} \\[1em] = -\dfrac{2}{4} \\[1em] = -\dfrac{1}{2}.$

Hence, cos² x° - sin² x° = $-\dfrac{1}{2}$.