How do you solve #-2sinx=-sinxcosx# for #0<=x<=2pi#?

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Answer 1 · 2016-10-23T21:10:33

There is no solution because it reduces to #cos(x) = 2#, which cannot be.

Answer 2 · 2016-10-23T21:14:47

#x in {0, pi, 2pi}#

#-2sin(x)=-sin(x)cos(x)#

#=> sin(x)cos(x)-2sin(x) = 0#

#=> sin(x)(cos(x)-2) = 0#

#=> sin(x) = 0 or cos(x)-2 = 0#

The second equation has no solutions, so we have

#sin(x) = 0#

In general, #sin(x) = 0 <=> x = pin, n in ZZ#.

On the interval #[0, 2pi]#, there are three such values, corresponding to #n=0, n=1, n=2#. Thus, our solution set is

#x in {0, pi, 2pi}#