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) \[x=2n\pi \pm {{\cos }^{-1}}(1/6)\]
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 =\[\frac{1}{3}\] \[\Rightarrow \,\,x=2n\pi \pm {{\cos }^{-1}}(1/3)\].
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