If sin2x-4 sin4x-4 sin2x cos2x-4-sin22x-4 sin2x-1-9 and 0 - x - - then the value of x is-

The correct options are B π6 D 5π 6 . sin^ 2 x+4sin^ 4 x 4sin^ 2 xcos^ 2 x 4 sin^ 2 2x 4sin^ 2 x #160;= 1 9 ⇒sin^ 2 x+4sin^ 4 x 4sin^ ...

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If sin2x+4 ...

Question

If $\frac{{sin}^{2} x + 4 {sin}^{4} x - 4 {sin}^{2} x {cos}^{2} x}{4 - {sin}^{2} 2 x - 4 {sin}^{2} x} = \frac{- 1}{9}$ and $0 < x <$ $π$ , then the value of $x$ is-

\frac{π}{3}

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\frac{π}{6}

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\frac{2 π}{3}

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\frac{5 π}{6}

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\frac{s i n^{2} 2 x + 4 s i n^{4} x - 4 s i n^{2} x c o s^{2} x}{4 - s i n^{2} 2 x - 4 s i n^{2} x} = \frac{1}{9}

π

x

\frac{{sin}^{2} 2 x + 4 {sin}^{4} x - 4 {sin}^{2} x {cos}^{2} x}{4 - {sin}^{2} 2 x - 4 {sin}^{2} x} = \frac{1}{9}

x

\frac{{sin}^{2} 2 x + 4 {sin}^{4} x - 4 {sin}^{2} x {cos}^{2} x}{4 - {sin}^{2} 2 x - 4 {sin}^{2} x} = \frac{1}{9}

^{'} a^{'}

^{'} b^{'}

f (x)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x + a \sqrt{2} sin x, & 0 \leq x < \frac{π}{4} 2 x cot x + b, & \frac{π}{4} \leq x \leq \frac{π}{2} a cos 2 x - b sin x, & \frac{π}{2} < x \leq π \end{matrix}

x = \frac{π}{4}, \frac{π}{2}

x \in (π, \frac{3 π}{2})

4 {cos}^{2} (\frac{π}{4} - \frac{x}{2}) + \sqrt{4 {sin}^{4} x + {sin}^{2} 2 x}

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