Question

The equation $3\sin^{2}x + 10\cos x - 6 = 0$ is satisfied if

• A. $x = n\pi \pm \cos^{- 1}\left( \frac{1}{3} \right)$
• B. $x = 2n\pi \pm \cos^{- 1}\left( \frac{1}{3} \right)$
• C. $x = n\pi \pm \cos^{- 1}\left( \frac{1}{6} \right)$
• D. $x = 2n\pi \pm \cos^{- 1}\left( \frac{1}{6} \right)$

Answer 1

Answer: (B)

$x = 2n\pi \pm \cos^{- 1}\left( \frac{1}{3} \right)$

Answer 2

$3\sin^{2}x + 10\cos x - 6 = 0$ ⇒ $3(1 - \cos^{2}x) + 10\cos x - 6 = 0$

on solving, $(\cos x - 3)(3\cos x - 1) = 0$. Either $\cos x = 3$

(which is not possible) or $\cos x = \frac{1}{3}$ $\mathbf{\Rightarrow}$ $\mathbf{x = 2n\pi}\mathbf{\pm}\mathbf{\cos}^{\mathbf{-}\mathbf{1}}\left( \frac{\mathbf{1}}{\mathbf{3}} \right)$.