How do you solve #2sinx-1=0#?

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Answer 1 · 2016-09-11T17:33:26

#x=npi+(-1)^npi/6#, where #n# is an integer.

As #2sinx-1=0#

#2sinx=1# or

#sinx=1/2=sin(pi/6)#

Now as sine function is positive in first and second quadrant

Hence #x=pi/6# or #x=pi-pi/6=(5pi)/6#

But as sine function has cycle of #2pi#

#x=2npi+pi/6# or #x=2npi+(5pi)/6#

We can write it in a single equation as #x=npi+(-1)^npi/6#, where #n# is an integer.