The solution of the equation sin x-cos x 1-sin 2x-2- - x- is

The correct option is C π4 ( cosx + sinx)^1 + sin2x = 2 ⇒ ( cosx + sinx)^(cosx + sinx)^2 = 2 ⇒1√(2) cosx + 1√(2) sinx = ± 1 ⇒ cos(x ...

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Fundamental Laws of Logarithms

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Question

The solution of the equation ${(sin x + cos x)}^{1 + sin 2 x} = 2, - π \leq x \leq π$ , is

\frac{π}{2}

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π

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\frac{π}{4}

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none of these

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\sqrt{1 + cos 2 x} = \sqrt{2} {sin}^{- 1} (sin x), - π \leq x \leq π

\sqrt{1 + cos 2 x} = \sqrt{2} {sin}^{- 1} (sin x)

x \in [- π, π]

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} x + a^{2} \sqrt{2} s i n x, & 0 \leq x < \frac{π}{4} x c o t x + b, & \frac{π}{4} \leq x < \frac{π}{2} b s i n 2 x - a c o s 2 x, & \frac{π}{2} \leq x \leq π \end{matrix}

[0, π]

^{'} 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}

A

0 < x < \frac{π}{2}

{sin}^{- 1} (c o s x) + {cos}^{- 1} (s i n x) = π - 2 x

{cos}^{- 1} x = \frac{π}{2} - {sin}^{- 1} x \forall x \in [0, 1]

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