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Question Number 74130 by mathocean1 last updated on 19/Nov/19

h e l l o] help me to solve it in] - Π; Π] \times] - Π; Π] please

{\begin{cases} x - y = \frac{Π}{6} \\ cosx - \sqrt{3} cosy = - \frac{1}{2} \end{cases}

Answered by Tanmay chaudhury last updated on 19/Nov/19

c o s (\frac{π}{6} + y) - \sqrt{3} c o s y = \frac{- 1}{2}

\frac{\sqrt{3}}{2} c o s y - \frac{1}{2} s i n y - \sqrt{3} c o s y = \frac{- 1}{2}

\frac{- 1}{2} s i n y - \frac{\sqrt{3}}{2} c o s y = \frac{- 1}{2}

\frac{1}{2} s i n y + \frac{\sqrt{3}}{2} c o s y = \frac{1}{2}

s i n (\frac{π}{3} + y) = s i n \frac{π}{6}

\frac{π}{3} + y = \frac{π}{6} \to y = - \frac{π}{6}

s o x = y + \frac{π}{6} \to x = 0

s i n (\frac{π}{3} + y) = s i n \frac{π}{6} = s i n (π - \frac{π}{6})

\frac{π}{3} + y = \frac{5 π}{6} \to y = \frac{5 π}{6} - \frac{2 π}{6} = \frac{π}{2}

x = \frac{π}{6} + y \to x = \frac{π}{6} + \frac{3 π}{6} = \frac{4 π}{6}