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Answer 1 · 2016-06-08T15:01:54

$x=pi/2+kpi,pi/4+(kpi)/2,k in ZZ$

Use the identity $sin2x=2sinxcosx$ to rewrite the expression.

$sin2xsinx=cosx" "=>" "2sinxcosx(sinx)=cosx$

$=>" "2sin^2xcosx=cosx$

Subtract $cosx$ from each side.

$2sin^2xcosx-cosx=0$

Factor $cosx$ .

$cosx(2sin^2x-1)=0$

Set both of these terms equal to $0$ .

$cosx=0$

This occurs at $x=pi/2$ and $x=(3pi)/2$ on the interval $[0,2pi)$ , but can be generalized as $x=pi/2+kpi$ where $k$ is an integer.

The other term is

$2sin^2x-1=0$

$sin^2x=1/2$

$sinx=+-sqrt(1/2)=+-1/sqrt2=+-sqrt2/2$

Sine equals $+-sqrt2/2$ at $x=pi/4,(3pi)/4,(5pi)/4,(7pi)/4$ on $[0,2pi)$ , which can be generalized to $pi/4+(kpi)/2$ , where $k$ is an integer.

Note that the mathematical way to express that $k$ is an integer is to say that $k in ZZ$ .