The solution set of sin⁡x+π4=sin⁡2x

sin⁡x+π4=sin⁡2x x+π4=nπ+(−1)n2x⇒x1−2(−1)n=nπ−π4⇒x=nπ−π41−2(−1)n,n∈Z

The solution set of sin⁡x+π4=sin⁡2x

sin⁡x+π4=sin⁡2x x+π4=nπ+(−1)n2x⇒x1−2(−1)n=nπ−π4⇒x=nπ−π41−2(−1)n,n∈Z

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The solution set of $\sin (x + \frac{π}{4}) = \sin 2 x$

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$\frac{nπ - (\frac{π}{4})}{1 + (- 1)^{n} (2)}$

$\frac{nπ + (\frac{π}{4})}{1 + (- 1)^{n} (2)}$

$\frac{nπ + (\frac{π}{4})}{1 - (- 1)^{n} (2)}$

$\frac{nπ - (\frac{π}{4})}{1 - (- 1)^{n} (2)}$

answer is D.

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Detailed Solution

$\begin{array}{l} \sin (x + \frac{π}{4}) = \sin 2 x [x + \frac{π}{4}] = nπ + (- 1)^{n} 2 x \\ \Rightarrow x (1 - 2 (- 1)^{n}) = nπ - \frac{π}{4} \\ \Rightarrow x = \frac{nπ - \frac{π}{4}}{1 - 2 (- 1)^{n}}, n \in Z \end{array}$

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The solution set of sin⁡x+π4=sin⁡2x

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The solution set of $\sin (x + \frac{π}{4}) = \sin 2 x$