Question

√(2)*sinX-√(2)*cosX=1

Answer 1

$\mathtt{\sqrt{2}sinx-\sqrt{2}cosx=1;~sinx-cosx=\frac{1}{\sqrt{2}};~}\\\\\mathtt{\sqrt{2}sin(x-\frac{\pi}{4})=\frac{\sqrt{2}}{2};~sin(x-\frac{\pi}{4})=\frac{1}{2};~\left[\begin{array}{ccc}\mathtt{x-\frac{\pi}{4}=\frac{\pi}{6}+2n\pi}\\\mathtt{x-\frac{\pi}{4}=\frac{2\pi}{3}+2n\pi}\end{array}\right}$

ОТВЕТ:  $\mathtt{\left[\begin{array}{ccc}\mathtt{x=\frac{5\pi}{12}+2n\pi}\\\\\mathtt{x=\frac{11\pi}{12}+2n\pi}\end{array}\right,~n\in Z}$

Answer 2

To solve the equation √(2) * sin(X) - √(2) * cos(X) = 1, we can follow these steps:

Step 1: Combine the like terms on the left side of the equation. Step 2: Isolate either sin(X) or cos(X) by moving the other term to the right side of the equation. Step 3: Use trigonometric identities to simplify the equation and solve for X.

Let's start with Step 1:

√(2) * sin(X) - √(2) * cos(X) = 1

Step 1: Combine like terms on the left side:

√(2) * (sin(X) - cos(X)) = 1

Step 2: Isolate sin(X) - cos(X):

(sin(X) - cos(X)) = 1 / √(2)

Step 3: Now we can use the trigonometric identity sin(π/4) = cos(π/4) = 1 / √(2):

sin(X) - cos(X) = sin(π/4)

Now, we'll use another trigonometric identity sin(θ - π/4) = sin(θ) * cos(π/4) - cos(θ) * sin(π/4):

sin(X - π/4) = sin(X) * cos(π/4) - cos(X) * sin(π/4)

sin(π/4) is 1/√(2), and cos(π/4) is also 1/√(2), so we can substitute these values:

sin(X - π/4) = (1/√(2)) * sin(X) - (1/√(2)) * cos(X)

Now, we know that sin(X) - cos(X) = sin(π/4), so we can make the substitution:

sin(X - π/4) = (1/√(2)) * sin(π/4)

Since sin(π/4) is 1/√(2), we have:

sin(X - π/4) = (1/√(2)) * (1/√(2))

sin(X - π/4) = 1/2

Now, to solve for X, we can take the inverse sine (arcsin) of both sides:

X - π/4 = arcsin(1/2)

X = arcsin(1/2) + π/4

Using the unit circle or a calculator, we find that arcsin(1/2) = π/6:

X = π/6 + π/4

X = (π + 3π)/12

X = 4π/12

X = π/3

So the solution to the equation √(2) * sin(X) - √(2) * cos(X) = 1 is X = π/3.