Og Function: -cos(3x)

Question

Derivative Function: 3 sin(3x)
The period for the function and its derivative is 2π/3 radians. Over one period, when is the derivative function equal to its original function? That is, when is f(x) = f '(x) over 0≤x≤2π/3? Round your final answer to three decimal places.

Reiny or anyone Please help me with step by step solution algebraically

Answer 1

So in effect we are solving
3 sin(3x) = -cos(3x)
sin(3x)/cos(3x) = -1/3
tan(3x) = -1/3 , but we know we are in quadrant II for the given domain, since the tangent is negative in quadrant II
set your calculator to RAD, find arctan (+1/3)
I get 3x = .32175
so 3x = .32175 OR 3x = π - .32175 = appr 2.819842..
x = .10725 OR x = .93995
remember that the period of tan is π, so by adding π to any
previous answer will yield a new answer, as long as we stay below 2π/3 (2.094...)