How do you solve #4sinx=2sinx+sqrt3#?

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Answer 1 · 2016-09-02T02:12:33

#x = 60^@#

#4sinx=2sinx+sqrt3#
#2sinx=sqrt3#
#sinx=sqrt3/2#

#x=60^@#

Answer 2 · 2016-09-04T07:53:09

#x = (pi) / (3)#

We have: #4 sin(x) = 2 sin(x) + sqrt(3)#

Let's subtract #2 sin(x)# from both sides of the equation:

#=> 2 sin(x) = sqrt(3)#

Then, we can divide both sides by #2# to get:

#=> sin(x)= (sqrt(3)) / (2)#

#=> x = (pi) / (3)#

Therefore, the solution to the equation is #x = (pi) / (3)#.