Answer:
x = n·π -arctan(1/ln(3)) . . . . n = 0, 1, 2
Explanation:
Critical points are where the derivative is zero or undefined. This function has no places where the derivative is undefined, so you need to find the values of x where the derivative is zero.
The product rule can be used to find the derivative of the function.
y = (3^x)sin(x)
y' = ln(3)(3^x)sin(x) +(3^x)cos(x) = (3^x)(ln(3)sin(x) +cos(x))
The factor 3^x is positive everywhere, so you will find y'=0 where ...
ln(3)sin(x) +cos(x) = 0
Dividing by cos(x) gives ...
ln(3)tan(x) +1 = 0
tan(x) = -1/ln(3)
x = -arctan(1/ln(3))
The tangent function is periodic with period π, so your solutions will have multiples of π added to this value of x:
x = n·π - arctan(1/ln(3)) . . . . . n = 0, 1, 2