How do you solve #2sinpix = 1# graphically and algebraically?

1 Answer
A. S. Adikesavan
Jul 17, 2018

#x = k +( - 1 )^k(1/6), k = 0, +-1, +-2, +-3,...#. I would improve this myself. Please do not edit. Another answer could be given, instead.

Explanation:

#sin pix = 1/2 = sin (pi/6)#. So,

#pix = kpi + ( - 1 )^k(pi/6)#

#rArr x = k +( - 1 )^k(1/6)#,

#k = 0, +-1, +-2, +-3,...#

Graphical solutions are approximations of rationals

#...-11/6, -7/6, 1/6, 5/6, 13/6, ...#
graph{(y-2 sin (3.141592654x) +1)((x+7/6)^2+y^2-0.01)((x-1/6)^2+y^2-.01)((x-5/6)^2+y^2-0.01)((x+11/6)^2+y^2-0.01)((x-13/6)^2+y^2-0.01)=0[-3 3 -3 2]}

Graph for solution near #- 5# is 5-sd #-5.1667#, by trial-and-error

root-bracketing method.

against #-31/6 = - 5.16666.....#:
graph{y-2 sin (3.141592654x) +1=0[-5.167 -5.166 -0.001 0.001]}
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