Question

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

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

Answer 1

Answer: (C), (D)

The correct options are
C $x=2n\pi +{\cos }^{-1}\left(\frac{1}{3}\right)$
D $x=2n\pi -{\cos }^{-1}\left(\frac{1}{3}\right)$

we have, $3{\sin }^{2}x+10\cos x-6=0$

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

$\Rightarrow 3{\cos }^{2}x-10\cos x+3=0$

$\Rightarrow (3\cos x-1)(\cos x-3)=0$

Therefore $\cos x=\frac{1}{3}$ (because $\cos x\ne 3$).

Hence $x=2n\pi \pm co{s}^{-1}\left(\frac{1}{3}\right)$.