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11. ⇒ (MHT CET 2021 20th September Evening Shift )

The principal solutions of $\sqrt{3} \sec x + 2 = 0$ are

A. $\frac{π}{6}, \frac{5 π}{6}$

B. $\frac{5 π}{6}, \frac{7 π}{6}$

C. $\frac{π}{3}, \frac{2 π}{3}$

D. $\frac{2 π}{3}, \frac{4 π}{3}$

Correct answer option is (B)

$\begin{aligned} \sqrt{3} \sec x + 2 = 0 \\ ∴ \sec x = \frac{- 2}{\sqrt{3}} \Rightarrow \cos x = \frac{- \sqrt{3}}{2} \\ ∴ \cos x = \cos (π - \frac{π}{6}) = \cos (π + \frac{π}{6}) \Rightarrow x = \frac{5 π}{6}, \frac{7 π}{6} \end{aligned}$

12. ⇒ (MHT CET 2021 20th September Morning Shift )

If $x \in (0, \frac{π}{2})$ and $x$ satisfies the equation $\sin x \cos x = \frac{1}{4}$ , then the values of $x$ are

A. $\frac{π}{12}, \frac{5 π}{12}$

B. $\frac{π}{8}, \frac{3 π}{8}$

C. $\frac{π}{8}, \frac{π}{4}$

D. $\frac{π}{6}, \frac{π}{12}$

Correct answer option is (A)

$\begin{aligned} \sin x \cos x = \frac{1}{4} \\ ∴ 2 \sin x \cos x = \frac{2}{4} = \frac{1}{2} \Rightarrow \sin 2 x = \frac{1}{2} = \sin 30^{\circ} \\ ∴ 2 x = 30^{\circ} = \frac{π}{6} \Rightarrow x = \frac{π}{12} \\ ∴ x = \frac{π}{2} \pm \frac{π}{12} \Rightarrow x = \frac{π}{2} + \frac{π}{12} or x = \frac{π}{2} - \frac{π}{12} \\ ∴ x = \frac{7 π}{12} or \frac{5 π}{12} \Rightarrow x = \frac{π}{12}, \frac{5 π}{12} . . . . [∵ x \in (0, \frac{π}{2})] \end{aligned}$

⇒Trigonometric Functions