Question

Sin x + cos x = -1.

Answer 1

Sinx + Cosx = - 1

Разделим обе части на √2 , получим :

$\frac{1}{\sqrt{2} }Sinx+\frac{1}{\sqrt{2} }Cosx=-\frac{1}{\sqrt{2}}\\\\Cos\frac{\pi }{4}Sinx+Sin\frac{\pi }{4}Cosx=-\frac{1}{\sqrt{2} }\\\\Sin(x+\frac{\pi }{4})=-\frac{1}{\sqrt{2} }\\\\x+\frac{\pi }{4}=(-1)^{n}arcSin(-\frac{1}{\sqrt{2}})+\pi n,n\in z\\\\x+\frac{\pi }{4}=(-1)^{n+1}\frac{\pi }{4}+\pi n,n\in z\\\\x=(-1)^{n+1}\frac{\pi }{4}-\frac{\pi }{4}+\pi n,n\in z$

Answer 2

Замена: x=t+π; sin(x+π)+cos(t+π)=-1; -sin t-cos t=-1; sin t+cos t=1. Напомню, что синус и косинус не могут принимать значения большие 1. Поэтому если sin t или cos t <0 (или оба), их сумма не может равняться 1.  Поэтому t обязан принадлежать первой четверти. Ясно, что t=2πn и  t=(π/2)+2πn  являются решениями. Докажем, что других решений в первой четверти нет. Достаточно (в силу периодичности синуса и косинуса) доказать, что нет решений при t∈(0;π/2). Но это очевидно, так как в этом случае sin t и cos t являются катетами прямоугольного треугольника с гипотенузой 1, а по неравенству треугольника сумма двух сторон треугольника всегда больше третьей.

Итак, мы нашли значения для t, остается написать ответ для x.

Ответ: π+2πn, n∈Z; (3π/2)+2πn, n∈Z

Answer 3

To solve the equation sin(x) + cos(x) = -1, we can use trigonometric identities and techniques. Let's go step by step:

Step 1: Rewrite sin(x) and cos(x) in terms of a single trigonometric function. We can use the identity sin^2(x) + cos^2(x) = 1 to express either sin(x) or cos(x) in terms of the other trigonometric function.

sin^2(x) + cos^2(x) = 1 => sin^2(x) = 1 - cos^2(x) => sin(x) = ±√(1 - cos^2(x))

Step 2: Substitute sin(x) in the original equation. sin(x) + cos(x) = -1 => ±√(1 - cos^2(x)) + cos(x) = -1

Step 3: Solve for cos(x). To isolate cos(x), we need to square both sides of the equation to remove the square root.

(±√(1 - cos^2(x)) + cos(x))^2 = (-1)^2 => (1 - cos^2(x)) + 2cos(x)√(1 - cos^2(x)) + cos^2(x) = 1

Simplify the equation:

1 - cos^2(x) + cos^2(x) + 2cos(x)√(1 - cos^2(x)) = 1

Now, the squared terms cancel out:

2cos(x)√(1 - cos^2(x)) = 0

Step 4: Solve for cos(x) again. To proceed further, we consider two separate cases for the equation:

Case 1: 2cos(x)√(1 - cos^2(x)) = 0 Set each term separately to zero:

a) 2cos(x) = 0 => cos(x) = 0

b) √(1 - cos^2(x)) = 0 => 1 - cos^2(x) = 0 => cos^2(x) = 1 => cos(x) = ±1

Therefore, for case 1, we have three possible solutions for cos(x): cos(x) = 0, cos(x) = 1, and cos(x) = -1.

Case 2: √(1 - cos^2(x)) = 0 (ignoring the other term 2cos(x) since it doesn't have a solution here)

This case leads to no real solutions since the square root of a non-negative number can't be zero.

Step 5: Find the corresponding values of sin(x) for each of the solutions obtained in step 4.

For the solutions obtained in case 1: a) cos(x) = 0 => sin(x) = ±√(1 - cos^2(x)) = ±√(1 - 0) = ±1

b) cos(x) = 1 => sin(x) = ±√(1 - cos^2(x)) = ±√(1 - 1) = ±√0 = 0

c) cos(x) = -1 => sin(x) = ±√(1 - cos^2(x)) = ±√(1 - 1) = ±√0 = 0

So, the solutions for sin(x) corresponding to each case are: a) sin(x) = ±1 b) sin(x) = 0 c) sin(x) = 0

Therefore, the complete set of solutions for the equation sin(x) + cos(x) = -1 is: x = π/2 + 2nπ, where n is an integer (for sin(x) = 1) x = 3π/2 + 2nπ, where n is an integer (for sin(x) = -1) x = 2nπ, where n is an integer (for sin(x) = 0)