If Cosxsin2x-32sin x+12=1 then x=

|cos x|sin2x-32 sin x+12=1.(cos x≠0) 2sin2x-3sin x+1=0 (2sinx-1) (sin x-1)=0 sin x=12, 1 x=π6, π2. x≠π2cos x=±1 x=0, π, 2π sin x=12=sinπ6 x=nπ+(-1)nπ6,∴ x=nπ+(-1)nπ6∪nπ.

If Cosxsin2x-32sin x+12=1 then x=

|cos x|sin2x-32 sin x+12=1.(cos x≠0) 2sin2x-3sin x+1=0 (2sinx-1) (sin x-1)=0 sin x=12, 1 x=π6, π2. x≠π2cos x=±1 x=0, π, 2π sin x=12=sinπ6 x=nπ+(-1)nπ6,∴ x=nπ+(-1)nπ6∪nπ.

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If ${|Cosx|}^{\sin^{2} x - \frac{3}{2} \sin x + \frac{1}{2}} = 1$ then $x =$

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$\{nπ + {(- 1)}^{n} \frac{π}{6}\} \cup \{nπ\}$

$(2 n + 1) \frac{π}{6}$

$nπ \pm \frac{π}{4}$

$2 nπ \pm \frac{π}{8}$

answer is A.

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Detailed Solution

$| \cos x |^{\sin^{2} x - \frac{3}{2} \sin x + \frac{1}{2}} = 1 . (\cos x \neq 0) 2 \sin^{2} x - 3 \sin x + 1 = 0 (2 sinx - 1) (\sin x - 1) = 0 \sin x = \frac{1}{2}, 1 x = \frac{π}{6}, \frac{π}{2} . (x \neq \frac{π}{2})$

$\cos x = \pm 1 x = 0, π, 2 π \sin x = \frac{1}{2} = \sin \frac{π}{6} x = nπ + {(- 1)}^{n} \frac{π}{6}, ∴ x = \{nπ + {(- 1)}^{n} \frac{π}{6}\} \cup \{nπ\} .$