Question

Система tgx+tgy=1 x+y=п/4​

Answer 1

Ответ:

(см. объяснение)

Объяснение:

$tg(x)+tg(y)=1\\x+y=\dfrac{\pi}{4}\\\\\dfrac{sin(x+y)}{cos(x)cos(y)}=1\\x+y=\dfrac{\pi}{4}\\\\\dfrac{\dfrac{\sqrt{2}}{2}}{cos(x)cos(y)}=1\\x+y=\dfrac{\pi}{4}\\\\\\\dfrac{cos(x-y)}{2}+\dfrac{cos(x+y)}{2}=\dfrac{\sqrt{2}}{2}\\x+y=\dfrac{\pi}{4}\\\\\\\dfrac{cos(x-y)}{2}+\dfrac{cos(\dfrac{\pi}{4})}{2}=\dfrac{\sqrt{2}}{2}\\x+y=\dfrac{\pi}{4}\\\\\\\dfrac{cos(x-y)}{2}=\dfrac{\sqrt{2}}{4}\\x+y=\dfrac{\pi}{4}\\\\\\cos(x-y)=\dfrac{\sqrt{2}}{2}\\x+y=\dfrac{\pi}{4}$

Найдем x-y:

$\left[\begin{array}{c}x-y=\dfrac{\pi}{4}+2k\pi,\; k\in Z\\x-y=\dfrac{7\pi}{4}+2k\pi,\; k\in Z\end{array}\right \\x+y=\dfrac{\pi}{4}\\\\1)\\x-y=\dfrac{\pi}{4}+2k\pi,\; k\in Z\\x+y=\dfrac{\pi}{4}\\\\\\2x=\dfrac{\pi}{2}+2k\pi,\; k\in Z\\x=\dfrac{\pi}{4}+k\pi,\; k\in Z\\y=-k\pi,\; k\in Z\\\\2)\\x-y=\dfrac{7\pi}{4}+2k\pi,\; k\in Z\\x+y=\dfrac{\pi}{4}\\\\\\x=\pi+k\pi,\; k\in Z\\y=-\dfrac{3\pi}{4}-k\pi, k\in Z$

Answer 2

To solve the system of equations, we'll use the substitution or elimination method. The given system is:

1. tg(x) + tg(y) = 1
2. x + y = π/4

Let's solve it using the substitution method:

Step 1: Solve the second equation (2) for y: y = π/4 - x

Step 2: Substitute the expression for y from step 1 into the first equation (1): tg(x) + tg(π/4 - x) = 1

Step 3: Now, we'll use the trigonometric identity: tg(π/4 - x) = 1 - tg(x). So, we have: tg(x) + (1 - tg(x)) = 1

Step 4: Simplify the equation: tg(x) + 1 - tg(x) = 1 1 = 1

Step 5: The equation 1 = 1 is an identity, which means it holds true for all values of x. This means that the system has infinitely many solutions.

In summary, the given system of equations tg(x) + tg(y) = 1 and x + y = π/4 has infinitely many solutions.