The solution set of sin⁡(x+π4)=sin⁡2x

[x+π4]=nπ+(−1)n2x⇒x(1−2(−1)n)=nπ−π4,⇒x=nπ−π41−2(−1)n,n∈Z

The solution set of sin⁡(x+π4)=sin⁡2x

[x+π4]=nπ+(−1)n2x⇒x(1−2(−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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$\frac{n π - (\frac{π}{4})}{1 + (- 1)^{n} (2)}$

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

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

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

answer is D.

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Detailed Solution

$[x + \frac{π}{4}] = n π + (- 1)^{n} 2 x$

$\Rightarrow x (1 - 2 (- 1)^{n}) = n π - \frac{π}{4}$ ,

$\Rightarrow x = \frac{n π - \frac{π}{4}}{1 - 2 (- 1)^{n}}, n \in Z$

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The solution set of sin⁡(x+π4)=sin⁡2x

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