Trigonometric Equations Question 8
Question: The equation $ 3{{\sin }^{2}}x+10\cos x-6=0 $ is satisfied, if
[UPSEAT 2001]
Options:
A) $ x=n\pi \pm {{\cos }^{-1}}(1/3) $
B) $ x=2n\pi \pm {{\cos }^{-1}}(1/3) $
C) $ x=n\pi \pm {{\cos }^{-1}}(1/6) $
D) $ x=2n\pi \pm {{\cos }^{-1}}(1/6) $
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Answer:
Correct Answer: B
Solution:
- $ 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} $
$ \Rightarrow x=2n\pi \pm {{\cos }^{-1}}(1/3) $ .
BETA
