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Question Number 106579 by john santu last updated on 06/Aug/20

G iven \cos x + \cos y = \frac{3}{2} + \cos (x + y)

where x, y \in [0, 2 π] . find x & y

Answered by bobhans last updated on 06/Aug/20

\cos x + \cos y = \frac{3}{2} + \cos xcos y - \sin xsin y

\cos x + \cos y - \cos xcos y + \sin xsin y = \frac{3}{2}

recall \to {\begin{cases} \cos x = \frac{1 - \tan^{2} (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} \\ \sin x = \frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} \end{cases}

set \tan (\frac{x}{2}) = p & \tan (\frac{y}{2}) = q

\Leftrightarrow \frac{1 - p^{2}}{1 + p^{2}} + \frac{1 - q^{2}}{1 + q^{2}} - (\frac{1 - p^{2}}{1 + p^{2}}) (\frac{1 - q^{2}}{1 + q^{2}}) + \frac{4 pq}{(1 + p^{2}) (1 + q^{2})} = \frac{3}{2}

{9 p}^{2} q^{2} - 8 pq + p^{2} + q^{2} + 1 = 0

{(3 pq - 1)}^{2} + {(p - q)}^{2} = 0

\to {\begin{cases} p = q \\ p = \pm \frac{1}{\sqrt{3}} \end{cases}

case (1) {\begin{cases} p = \frac{1}{\sqrt{3}} \Rightarrow \tan (\frac{x}{2}) = \tan (\frac{π}{6}) \\ q = \frac{1}{\sqrt{3}} \Rightarrow \tan (\frac{y}{2}) = \tan (\frac{π}{6}) \end{cases}

\to {\begin{cases} x = \frac{π}{3} \\ y = \frac{π}{3} \end{cases} \Rightarrow (\frac{π}{3}, \frac{π}{3})

case (2)

\to {\begin{cases} p = - \frac{1}{\sqrt{3}} \Rightarrow \tan (\frac{x}{2}) = \tan (- \frac{π}{6}) \\ q = - \frac{1}{\sqrt{3}} \Rightarrow \tan (\frac{y}{2}) = \tan (- \frac{π}{6}) \end{cases}

\to {\begin{cases} x = \frac{5 π}{3} \\ y = \frac{5 π}{3} \end{cases} \Rightarrow (\frac{5 π}{3}, \frac{5 π}{3})

Commented by john santu last updated on 06/Aug/20

cool \dots & jooss ♠