Question 1081023
this function is periodic in {{{x}}} with period {{{30}}}(ie one wavelength)

so, the "primary interval for this equation"  is {{{30}}} units 

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to calculate where it starts on {{{x}}} axis, we have

{{{80sin((x*pi/15) - pi/5) + 340}}} where {{{A=80}}},{{{B=pi/15}}},{{{C=pi/5}}},and {{{D=340}}}



{{{0<= Bt+C <= 2pi}}} for any sinusoidal function

{{{0 <= (pi/15)t+pi/5 <=2pi}}}

{{{0-pi/5<=(pi/15)t+pi/5-pi/5<=2pi-pi/5}}}

{{{0-pi/5<=(pi/15)t<=10pi/5-pi/5}}}

{{{0-pi/5<=(pi/15)t<= 9pi/5}}}

{{{-pi/5<=(pi/15)t<=9pi/5}}}


{{{-(pi/5)/(pi/15)<=(pi/15)t /(pi/15)<=(9pi/5)/(pi/15)}}}

{{{-3<=t <=(9*15pi)/(5pi)}}}

{{{-3<=t <=9*3}}}

{{{highlight(-3<=t <=27)}}}-> primary period


so, as you can see, the distance between {{{-3}}} and {{{27}}} on x-axis is {{{30}}} units