3sinx+7cosx=Rsinx+a R0

To solve this equation, we can use the trigonometric identities:

sin(a + b) = sin(a)cos(b) + cos(a)sin(b) cos(a + b) = cos(a)cos(b) - sin(a)sin(b)

Comparing the given equation with the right side of the equation using the above identities, we can equate the coefficients:

3sin(x) + 7cos(x) = Rsin(x + a)

Comparing the coefficients of sin(x) and cos(x) on both sides of the equation, we get:

3 = Rcos(a) 7 = Rsin(a)

Dividing these two equations, we have:

3/7 = (Rcos(a))/(Rsin(a))

Simplifying, we get:

3/7 = cot(a)

Taking the inverse cotangent of both sides, we have:

a = arccot(3/7)

Therefore, the solution to the equation 3sin(x) + 7cos(x) = Rsin(x + a) with R > 0 is x = x + a, where a = arccot(3/7)