If x+y=2π3 and sinxsiny=2 then

x+y=2π3⇒y=2π3−x∴sinx=2sin2π3−x = 232cosx+12sinx = 3cosx+sinx⇒cosx=0⇒x=nπ+π2⇒y=2π3−nπ−π2=π6−nπHence,x∈0,4π,x=π2,3π2,5π2,7π2and for x∈0,4π,y=π6,7π6,13π6,19π6

If x+y=2π3 and sinxsiny=2 then

x+y=2π3⇒y=2π3−x∴sinx=2sin2π3−x = 232cosx+12sinx = 3cosx+sinx⇒cosx=0⇒x=nπ+π2⇒y=2π3−nπ−π2=π6−nπHence,x∈0,4π,x=π2,3π2,5π2,7π2and for x∈0,4π,y=π6,7π6,13π6,19π6

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If $x + y = \frac{2 π}{3} a n d \frac{sinx}{\sin y} = 2$ then

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The number of values of $x \in$ $[0, 4 π]$ are 4

The number of values of $x \in [0, 4 π]$ are 2

The number of values of $y \in [0, 4 π]$ are 4

The number of values of $y \in [0, 4 π]$ are 8

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Detailed Solution

$\begin{array}{l} x + y = \frac{2 π}{3} \Rightarrow y = \frac{2 π}{3} - x \\ ∴ \sin x = 2 \sin (\frac{2 π}{3} - x) \\ = 2 [(\frac{\sqrt{3}}{2}) \cos x + \frac{1}{2} \sin x] \\ = \sqrt{3} \cos x + \sin x \\ \Rightarrow \cos x = 0 \Rightarrow x = n π + \frac{π}{2} \\ \Rightarrow y = \frac{2 π}{3} - n π - \frac{π}{2} = \frac{π}{6} - n π \\ H e n c e, \\ x \in [0, 4 π], x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \frac{7 π}{2} \\ a n d f o r x \in [0, 4 π], y = \frac{π}{6}, \frac{7 π}{6}, \frac{13 π}{6}, \frac{19 π}{6} \end{array}$