Question

Решите систему: cos(x)*sin(y)=1/4 sin(x)*cos(y)=3/4

Answer 1

Решение:

Во-первых, необходимо, следуя формуле $\left \{ {{cos(x)sin(y)=a} \atop {sin(x)cos(y) = b}}$, сложим и вычтим уравнения системы и получим равносильную систему.

$\left \{ {{sin(x-y)=\cfrac{3-1}{4}}} \atop {sin(x+y) = \cfrac{1+3}{4}} \right.\Rightarrow \left \{ {{sin(x-y)=\cfrac{2}{4}\rightarrow \cfrac{1}{2}}} \atop {sin(x+y) = \cfrac{4}{4}\rightarrow1}\left \right.$

Теперь по формуле синусов рассмотрим систему в более банальной форме.

$\left \{ {{sin(x-y)=\cfrac{1}{2}} \atop {sin(x+y)=1}} \right. \Rightarrow \left \{ {{x-y=(-1)^k \cfrac{\pi}{6}+\pi k} \atop {x+y=\cfrac{\pi}{2}+2\pi n}}\right.$

Теперь сложим первое и второе уравнения и получим далее по алгоритму результат.

$2x=\cfrac{\pi}{2}+(-1)^k \cfrac{\pi}{6} +2\pi n + \pi k \Rightarrow \\ \\ \boxed{x}=\cfrac{\pi}{4}+(-1)^k \cfrac{\pi}{12}+\pi n + \cfrac{\pi}{2} k \\ \\ \boxed{y}=\cfrac{\pi}{4}-(-1)^k \cfrac{\pi}{12}+\pi n - \cfrac{\pi}{2} k$

Ответ:   $\boxed{\bf x=\cfrac{\pi}{4}+(-1)^k \cfrac{\pi}{12}+\pi n + \cfrac{\pi}{2} k; \: \: y=\cfrac{\pi}{4}-(-1)^k \cfrac{\pi}{12}+\pi n - \cfrac{\pi}{2} k}$

Answer 2

To solve the given system of equations:

1. cos(x) * sin(y) = 1/4
2. sin(x) * cos(y) = 3/4

We can use the trigonometric identity: sin(2θ) = 2 * sin(θ) * cos(θ). Let's rewrite the second equation using this identity:

sin(x) * cos(y) = 3/4 2 * sin(x) * cos(y) = 3/2

Now, let's add the two equations together:

cos(x) * sin(y) + 2 * sin(x) * cos(y) = 1/4 + 3/2 cos(x) * sin(y) + 2 * sin(x) * cos(y) = 2/4 + 6/4 cos(x) * sin(y) + 2 * sin(x) * cos(y) = 8/4 cos(x) * sin(y) + 2 * sin(x) * cos(y) = 2

Now, we can use the sum-to-product identities:

sin(a) * cos(b) + cos(a) * sin(b) = sin(a + b)

Applying this to our equation:

sin(x + y) = 2

Now, to find the value of (x + y), we need to take the arcsin of both sides:

x + y = arcsin(2)

However, the arcsin function is only defined for values between -1 and 1, and since 2 is outside this range, the system has no real solution.

Therefore, the given system of equations has no real solution.