How do you solve #cos(x-pi/2)=sqrt3/2# for #-pi<x<pi#?

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Answer 1 · 2018-04-15T09:19:15

#color(blue)(pi/3,(2pi)/3)#

Identity:

#color(red)bb(cos(A-B)=cosAcosB+sinAsinB)#

#cos(x-pi/2)=cos(x)cos(pi/2)+sin(x)sin(pi/2)#

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \=cosx(0)+sinx(1)#

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \=sinx#

#:.#

#sinx=sqrt(3)/2#

#x=arcsin(sinx)=arcsin(sqrt(3)/2)=>x=pi/3,(2pi)/3#

#color(blue)(pi/3,(2pi)/3)#