sinx-cosx-1-sinx-cosy-1-x-

Question Number 205219 by hardmath last updated on 13/Mar/24

{\begin{cases} sinx + cosx = 1 \\ sinx - cosy = 1 \end{cases}

\Rightarrow x = ?

Answered by Rasheed.Sindhi last updated on 13/Mar/24

{\begin{cases} sinx + cosx = 1 \dots i \\ sinx - cosy = 1 \dots i i \end{cases}; x = ?

\cos x + \cos y = 0

\cos y = - \cos x

i i \Rightarrow \sin x + \cos x = 1

\sin^{2} x + 2 \sin x \cos x + \cos^{2} x = 1

1 + 2 \sin x \cos x = 1

\sin x \cos x = 0

\sin x = 0 ∣ \cos x = 0

x = n π ∣ x = \frac{π}{2} + n π

Commented by mr W last updated on 13/Mar/24

t h r o u g h s q u a r i n g y o u a d d e d a l s o

t h e r o o t s f o r e q n . \sin x + \cos x = - 1 .

i t h i n k o n l y x = 2 n π o r 2 n π + \frac{π}{2} a r e

s u i t a b l e .

Commented by Rasheed.Sindhi last updated on 13/Mar/24

V e r y r i g h t s i r! T h a n k s .

Answered by MM42 last updated on 13/Mar/24

\sqrt{2} s i n (x + \frac{π}{4}) = 1 \Rightarrow {\begin{cases} x = 2 k π \\ x = 2 k π + \frac{π}{2} \end{cases}

i f x = 2 k π \Rightarrow c o s y = - 1 \Rightarrow y = (2 k + 1) π

i f x = 2 k π + \frac{π}{2} \Rightarrow c o s y = 0 \Rightarrow y = k π + \frac{π}{2}

\Rightarrow A n s = {(x, y) ∣ x = 2 k π & y = (2 k π + 1) π o r x = 2 k π + \frac{π}{2} & y = k π + \frac{π}{2}}

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