Question

Sinx +cos x = корень из 2

Answer 1

$sinx + cosx= \sqrt{2} \\ \frac{1}{ \sqrt{2} } cosx - \frac{1}{ \sqrt{2} }x=-1 \\ cos \frac{\pi}{4}cosx-sin \frac{\pi}{4}sinx=-1 \\ cos(\frac{\pi}{4}+x)=-1 \\ \frac{\pi}{4} +x=arccos(-1)+2\pi n \ \ \ \ \ \boxed{n \in Z} \\ \frac{\pi}{4} +x = \pi + 2\pi n \ \ \ \ \ \boxed{n \in Z} \\ x = \frac{3 \pi}{4} + 2\pi n \ \ \ \ \ \boxed{n \in Z}$

Answer 2

To solve the equation sin(x) + cos(x) = √2, we can use trigonometric identities. The equation can be rewritten as:

sin(x) + cos(x) = √2

Now, let's square both sides of the equation to eliminate the square root:

(sin(x) + cos(x))^2 = (√2)^2

sin^2(x) + 2sin(x)cos(x) + cos^2(x) = 2

Using the trigonometric identity sin^2(x) + cos^2(x) = 1, the equation becomes:

1 + 2sin(x)cos(x) = 2

Now, rearrange the terms:

2sin(x)cos(x) = 2 - 1

2sin(x)cos(x) = 1

Now, we can use the double angle formula for sine, which states that sin(2x) = 2sin(x)cos(x):

sin(2x) = 1

To solve for x, we need to find the values of 2x. We know that sin(π/4) = 1, so:

2x = π/4 + 2nπ or 2x = 3π/4 + 2nπ

where n is an integer.

Now, to find the values of x, divide both sides of the equations by 2:

x = π/8 + nπ or x = 3π/8 + nπ

where n is an integer.

These are the general solutions for the equation sin(x) + cos(x) = √2. It has infinitely many solutions due to the periodicity of trigonometric functions. If you need specific values, you can substitute different integers for n and calculate the corresponding values of x.