Question

Sin3xcos(x+П/4)+cos3xsin(x+а/4)=0 СРОЧНО!!помогите решить пожалуйста

Answer 1

Sin(x)+sin(3x)+cos(x)+cos(3x)=0[sin(x)+sin(3x)]+[cos(x)+cos(3x)]=02sin((3x+x)/2)*cos((3x-x)/2)+2cos((3x+x)/2)*cos((3x-x)/2)=02sin(2x)*cos(x)+2cos(2x)*cos(x)=02*cos(x)*(sin(2x)+cos(2x))=01)
cos(x)=0x=pi/2+pi*n(2)sin(2x)+cos(2x)=0sin(2x)=-cos(2x)sin(2x)/cos(2x)=-cos(2x)/cos(2x)tg(2x)=-12x=arctg(-1)+pi*n2x-3*pi/4+pi*nx=3pi/8+pi*n/2

Answer 2

Solving the Equation Sin(3x)cos(x+π/4) + cos(3x)sin(x+α/4) = 0

To solve the equation Sin(3x)cos(x+π/4) + cos(3x)sin(x+α/4) = 0, we can use trigonometric identities and algebraic manipulations. Let's break down the steps:

Step 1: Apply the product-to-sum identities to simplify the equation. - The product-to-sum identities state that sin(A)cos(B) = (1/2)[sin(A+B) + sin(A-B)] and cos(A)sin(B) = (1/2)[sin(A+B) - sin(A-B)].

Applying the product-to-sum identities to the equation, we get: (1/2)[sin(4x + α/4) + sin(2x + α/2)] + (1/2)[sin(4x + α/4) - sin(2x + α/2)] = 0.

Simplifying further, we have: sin(4x + α/4) + sin(2x + α/2) + sin(4x + α/4) - sin(2x + α/2) = 0.

Step 2: Combine like terms. sin(4x + α/4) + sin(4x + α/4) + sin(2x + α/2) - sin(2x + α/2) = 0.

Simplifying further, we get: 2sin(4x + α/4) = 0.

Step 3: Solve for x. To solve for x, we need to find the values of 4x + α/4 that satisfy the equation 2sin(4x + α/4) = 0.

Using the property of sine function that sin(x) = 0 when x is an integer multiple of π, we have: 4x + α/4 = nπ, where n is an integer.

Solving for x, we get: x = (nπ - α/4)/4.

Therefore, the general solution for the equation Sin(3x)cos(x+π/4) + cos(3x)sin(x+α/4) = 0 is: x = (nπ - α/4)/4, where n is an integer.

Please note that the value of α/4 is not provided in the question, so you will need to substitute the appropriate value to find the specific solutions.

I hope this helps! Let me know if you have any further questions.