If cos2 2x - 2cos2 x - 1 - x - - - - - then x can take the values -

Solve the equation: $\sqrt{(\frac{1}{16} + c o s^{4} x - \frac{1}{2} c o s^{2} x)} + \sqrt{(\frac{9}{16} + c o s^{4} x - \frac{3}{2} c o s^{2} x)} = \frac{1}{2}$

Q. Find the set of value of

x

satisfying the equality

sin (x - \frac{π}{4}) - cos (x - \frac{π}{4}) = 1

and the inequality

\frac{2 cos 7 x}{cos 3 + sin 3} > 2^{cos 2 x}

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If cos22x+2cos2x=1 , x ∈ (π , π) , then x can take the values -

The correct option is B π4Given, cos22x+2cos2x=1⇒2cos2x−1+cos2x=1⇒4cos2x=2⇒cos2x=12⇒cosx=±1√2∴x=π4

If $c o s^{2} 2 x + 2 c o s^{2} x = 1$ , $x$ $\in$ ( $π$ , $π$ ) , then $x$ can take the values -

The correct option is B $\frac{π}{4}$
Given,

${cos}^{2} 2 x + 2 {cos}^{2} x = 1$

$\Rightarrow 2 {cos}^{2} x - 1 + {cos}^{2} x = 1$

$\Rightarrow 4 {cos}^{2} x = 2$

$\Rightarrow {cos}^{2} x = \frac{1}{2}$

$\Rightarrow cos x = \pm \frac{1}{\sqrt{2}}$

$∴ x = \frac{π}{4}$