Question

Sqrt2 sin(3pi/2-x) cos(3pi/2+x)=cosx

Answer 1

1) -корень из 2cosx sinx=cosx
-корень из 2cosxsinx-cosx=0
cosx(-корень из 2sinx+1)=0
cosx=0          или  sinx=1/корень из 2
x=Пk                      x= (-1) в степени к, 1/корень из 2 + 2Пк

Answer 2

To solve the equation sqrt(2) sin(3pi/2 - x) cos(3pi/2 + x) = cos(x), we can simplify and manipulate the expression using trigonometric identities.

Let's start by expanding the trigonometric functions:

sqrt(2) sin(3pi/2 - x) cos(3pi/2 + x) = cos(x)

sqrt(2) [sin(3pi/2) cos(x) - cos(3pi/2) sin(x)] = cos(x)

sqrt(2) [-1 * cos(x) - 0 * sin(x)] = cos(x)

• sqrt(2) cos(x) = cos(x)

Now, we can solve for cos(x):

• sqrt(2) cos(x) = cos(x)

(- sqrt(2) - 1) cos(x) = 0

Since cos(x) cannot be zero (as it is the denominator), we have:

• sqrt(2) - 1 = 0

sqrt(2) = 1

However, this is not a valid solution since sqrt(2) is not equal to 1.

Therefore, the given equation has no solution.