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Trigonometrical Equations
medium
The equation $3{\sin ^2}x + 10\cos x - 6 = 0$ is satisfied, if
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
$x = 2n\pi \pm {\cos ^{ - 1}}(1/6)$
Solution
(b) $3{\sin ^2}x + 10\cos x – 6 = 0$
$3(1 – {\cos ^2}x) + 10\cos x – 6 = 0$
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)$.
Standard 11
Mathematics
