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

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

How do you solve -2sinx=-sinxcosx-2sinx=-sinxcosx for 0<=x<=2pi0<=x<=2pi?