My attempt:
I tried to find the values of $\sin (x+y)$ and $\sin (x-y)$, but am unsure if this is the right approach and what conditions are to be fulfilled for the given system of equations to have solutions.
Given $$\sin x \cos y = 2\lambda – 4 \implies$$ $$2\sin x \cos y = 4\lambda-8 \implies$$ $$\sin(x+y) +\sin(x-y) =4\lambda – 8 \tag{1}$$
and
$$\sin y \cos x = 1-\lambda \implies$$ $$2\sin y \cos x = 2 – 2 \lambda\implies$$ $$\sin(x+y) +\sin(y-x) = – 2\lambda + 2\tag{2}$$
$\sin (y-x) = – \sin(x-y)$ and from $(1)$ and $(2)$
$$\sin(x+y) = \lambda – 3$$ $$\sin (x-y) =3\lambda – 5$$
What to do next (if I am on the right track)? Hints would be appreciated.
Best Answer
As you wrote,
$\sin(x+y) +\sin(x-y) =4\lambda - 8 \tag{1}$
$\sin(x+y) - \sin(x-y) = - 2\lambda + 2\tag{2}$
Now with $(1) + 2 \times (2)$
$3 \sin (x+y) - \sin (x-y) = -4$
What does it mean for values of $\sin (x+y)$ and $\sin (x-y)$ given value of $\sin$ function is in $(-1, 1)$?
That should lead you to the value of $\lambda$.