MCQ
The equation $3{\sin ^2}x + 10\cos x - 6 = 0$ is satisfied, if
- A$x = n\pi \pm {\cos ^{ - 1}}(1/3)$
- ✓$x = 2n\pi \pm {\cos ^{ - 1}}(1/3)$
- C$x = n\pi \pm {\cos ^{ - 1}}(1/6)$
- D$x = 2n\pi \pm {\cos ^{ - 1}}(1/6)$
$3(1 - {\cos ^2}x) + 10\cos x - 6 = 0$
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)$.
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x
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1
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a
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$a^2$
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....
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$a^n$
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f
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$^nC_0$
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$^nC_1$
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$^nC_2$
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....
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$^nC_2$
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