To solve the given equation sin(3x)cos(x) - cos(3x)sin(x) = √3/2, we can use the sum-to-product trigonometric identities.
Recall that sin(A + B) = sinAcosB + cosAsinB and cos(A + B) = cosAcosB - sinAsinB.
Let A = 3x and B = x. Then, sin(3x + x) = sin(4x) = sin(3x)cos(x) + cos(3x)sin(x) and cos(3x + x) = cos(4x) = cos(3x)cos(x) - sin(3x)sin(x).
Therefore, the equation becomes sin(4x) = √3/2.
We know that sin(π/3) = √3/2, so we can rewrite the equation as sin(4x) = sin(π/3).
Since sin has period 2π, the general solution for sin(4x) = sin(π/3) is:
4x = π/3 + 2nπ or 4x = π - π/3 + 2nπ, where n is an integer.
This simplifies to:
x = π/12 + nπ/2 or x = 5π/12 + nπ/2, where n is an integer.
Therefore, the solutions to the given equation are x = π/12 + nπ/2 or x = 5π/12 + nπ/2, where n is an integer.
To solve the given equation sin(3x)cos(x) - cos(3x)sin(x) = √3/2, we can use the sum-to-product trigonometric identities.
Recall that sin(A + B) = sinAcosB + cosAsinB and cos(A + B) = cosAcosB - sinAsinB.
Let A = 3x and B = x. Then, sin(3x + x) = sin(4x) = sin(3x)cos(x) + cos(3x)sin(x) and cos(3x + x) = cos(4x) = cos(3x)cos(x) - sin(3x)sin(x).
Therefore, the equation becomes sin(4x) = √3/2.
We know that sin(π/3) = √3/2, so we can rewrite the equation as sin(4x) = sin(π/3).
Since sin has period 2π, the general solution for sin(4x) = sin(π/3) is:
4x = π/3 + 2nπ or 4x = π - π/3 + 2nπ, where n is an integer.
This simplifies to:
x = π/12 + nπ/2 or x = 5π/12 + nπ/2, where n is an integer.
Therefore, the solutions to the given equation are x = π/12 + nπ/2 or x = 5π/12 + nπ/2, where n is an integer.