Solve-for-x-and-y-x-2-2x-sin-xy-1-0-

Question Number 15298 by Tinkutara last updated on 09/Jun/17

Solve for x and y

x^{2} + 2 x \sin (x y) + 1 = 0

Commented by prakash jain last updated on 10/Jun/17

x y = u

x^{2} + 2 x \sin (u) + 1 = 0

\sin u = - \frac{1 + x^{2}}{2 x}

f (x) = - \frac{1 + x^{2}}{2 x}

f (x) ⩽ - 1 o r f (x) ⩾ 1

o n l y 2 s o l u t i o n f o r x f o r f (x) = \pm 1

f (x) = - 1 \Rightarrow x = 1

f (x) = 1 \Rightarrow x = - 1

x = 1

\sin x y = - \frac{1 + x^{2}}{2 x} = - 1

\sin (y) = - 1 \Rightarrow y = - \frac{π}{2}

x = - 1

\sin (- y) = - \frac{1 + x^{2}}{2 x}

\sin (- y) = 1

\sin (y) = - 1 \Rightarrow \sin y = - 1 \Rightarrow y = - \frac{π}{2}

solutions :

x = 1, y = - \frac{π}{2} (y c a n t a k e 2 n π + \frac{3 π}{2})

x = - 1, y = - \frac{π}{2} (y c a n t a k e 2 n π + \frac{3 π}{2})

Commented by Tinkutara last updated on 10/Jun/17

But answer is (x, y) =

(1, (2 n + 1) \frac{3 π}{2}) and (- 1, (2 n + 1) \frac{π}{2})

Commented by prakash jain last updated on 10/Jun/17

x^{2} + 2 x \sin (x y) + 1 = 0

p u t x = 1

2 \sin y + 2 = 0

\sin y = - 1

w h a t i s t h e s o l u t i o n f o r y .

c o m p a r e i t w i t h u r {b o o k}^{'} s a n s w e r .

Commented by prakash jain last updated on 10/Jun/17

3 n π + \frac{3 π}{2}

n = 0 y = \frac{3 π}{2}

n = 1 y = 3 π + \frac{3 π}{2} = 3 π + π + \frac{π}{2} = 4 π + \frac{π}{2}

p u t x = 1, y = 4 π + \frac{π}{2} in original equation

does it satisfy the original equation ?

Answered by Tinkutara last updated on 09/Jul/17

Given equation can be written as

{[x + \sin (x y)]}^{2} + \cos^{2} (x y) = 0

∴ \cos (x y) = 0 and x + \sin (x y) = 0 .

\Rightarrow x = \pm 1

\cos y = 0; \sin y = - 1 \Rightarrow y = (2 n + 1) \frac{3 π}{2}

\Rightarrow (x, y) = (\pm 1, (2 n + 1) \frac{3 π}{2})