Question

I posted this question before but accidentally typed in the wrong values! Please help, I just need to know the basic formula my book didn't provide any examples.

For the function y = -2sin(x-(pi/3)) between x = 0 and x = 2pi : (6 marks)

For what value(s) of x does y have its maximum value?
For what value(s) of x does y have its minimum value?
For what value(s) of x does y=1?

Answer 1

dy/dx = -2cos(x - π/3)
at a max/min , dy/dx = 0
-2cos(x-π/3) = 0
cos(x-π/3) = 0
I know that cos π/2 = 0 or cos 3π/2 = 0
so x - π/3 = π/2 or x - π/3 = 3π/2
x = 5π/6 or x = 11π/6 , (150° or 330°)

when x = 5π/6 , y = -2sin(π/2) = -2
when x = 11π/6 , y = -2sin(3π/2) = +2

so the function has a maximum value of 2 when x = 11π/3
and a minimum value of -2 when x = 5π/6

when y = 1
-2sin(x-π/3) = 1
sin(x-π/3) = -1/2
I know sin 210° = -1/2 and sin 330°=-1/2
which is 7π/6 or 11π/6
x-π/3 = 7π/6 or x-π/3=11π/6
<b>x = 3π/2 or x = 13π/6</b>