2sin-2x-y-sin-y-cos-2x-sin-2x-sin-2y-2-Find-the-solution-

Question Number 197437 by dimentri last updated on 17/Sep/23

{\begin{cases} 2 \sin (2 x + y) \sin y = \cos 2 x \\ \sin 2 x - \sin 2 y = \sqrt{2} \end{cases}

F i n d t h e s o l u t i o n

Answered by Frix last updated on 17/Sep/23

x = \tan^{- 1} u \land y = \tan^{- 1} v

The equations are then easy to solve

\Rightarrow

x = n_{1} π + \frac{3 π}{8} \land y = n_{2} π - \frac{π}{8}

\lor

x = n_{1} π + \frac{π}{8} \land y = n_{2} π - \frac{3 π}{8}

For 0 ⩽ x, y < π

x = \frac{3 π}{8} \land y = \frac{7 π}{8} \lor x = \frac{π}{8} \land y = \frac{5 π}{8}

Answered by cortano12 last updated on 17/Sep/23

solution is

x = {\frac{π}{8} + k . π; \frac{3 π}{8} + ℓ . π}

y = {- \frac{π}{8} + k . π; \frac{5 π}{8} + ℓ . π}

Answered by MM42 last updated on 17/Sep/23

c o s 2 x - c o s 2 (x + y) = c o s 2 x \Rightarrow 2 (x + y) = k π + \frac{π}{2}

2 {c o s}^{2} (x + y) - 1 = 0 \Rightarrow c o s (x + y) = \pm \frac{\sqrt{2}}{2}

\Rightarrow x + y = k π + \frac{π}{4}; \dots, - \frac{3 π}{4}, \frac{π}{4}, \frac{5 π}{4}, \dots

2 s i n (x - y) c o s (x + y) = \sqrt{2}

\Rightarrow s i n (x - y) = \pm 1

\Rightarrow x - y = k π + \frac{π}{2}; - \frac{π}{2}, \frac{π}{2}, \frac{3 π}{2}, \dots

\Rightarrow x = k π + \frac{3 π}{8} & y = k π - \frac{π}{8}

Answered by witcher3 last updated on 17/Sep/23

2 \sin (2 x + y) \sin (y)

\cos (a) - \cos (b) = - 2 \sin (\frac{a - b}{2}) \sin (\frac{a + b}{2})

\Rightarrow 2 sn (2 x + y) \sin (y) = \cos (2 x) - \cos (2 x + 2 y) = \cos (2 x)

\Rightarrow \cos (2 x + 2 y) = 0 \Rightarrow 2 x + 2 y = \frac{π}{2} + n π

\sin (2 x) - \sin (2 y) = \sqrt{2}

\Rightarrow \sin (\frac{π}{2} + n π - 2 y) - \sin (2 y) = \sqrt{2}

\Rightarrow \cos (n π - 2 y) - \sin (2 y) = \sqrt{2}

\Rightarrow {(- 1)}^{n} \cos (2 y) - \sin (2 y) = \sqrt{2}

n = 2 k \Rightarrow

\cos (2 y) - \sin (2 y) = \sqrt{2}

\Rightarrow \cos (2 y + \frac{π}{4}) = 1

\Rightarrow 2 y + \frac{π}{4} = 2 k π \Rightarrow y = - \frac{π}{8} + m π, x = \frac{3 π}{8} + k π

n = 2 k + 1

2 y - \frac{π}{4} = π + 2 k π

\Rightarrow y = \frac{5 π}{8} + k π

x = - \frac{π}{8} + n π

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