How to solve for #x# in #sin(2x-pi/3)=1/2#?

Question

how to solve sin(2x-Pi/3)=1/2 ?

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Answer 1 · 2018-01-31T15:38:10

#x = pi/4#

This might look intimidating, but we'll take this one step at a
time

#sin(2x-pi/3) = 1/2#

#2x-pi/3 = sin^-1(1/2)#

#2x - pi/3 = pi/6#

#2x = pi/6 + pi/3#

we need a common denominator

#2x = pi/6 + pi/3 xx 2/2#

#2x = (pi+2pi)/6#

#2x = (3pi)/6#

simplify

#2x = pi/2#

#x = (pi/2)/2#

#x = pi/2 xx 1/2#

#x = pi/4#

To check our work, let's graph the equation

graph{y +1/2= sin(2x-pi/3)}

Yes, there is an #x#-intercept at #(pi/4,0)# so we were right!