sin2xcosx-sinx=0 interval(0,2pi)

Answer 1

Use double angle formula to reduce sin(2x) to 2sin(x)cos(x):
2sin(x)cos(x)cos(x)-sin(x)=0
sin(x)(2cos²(x)-1)=0
=>
sin(x)=0, or cos(x)=±1/√2
=>
Note: the interval was probably meant to be: [0,2π)
x=0, x=π for sin(x)=0
or
x=&pi/4, x=3π/4 for cos(x)=1/√2,
x=5π/4 or x=7π4 for cos(x)=-1/√2.

Answer 2

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Answer 3

Afrim, please do not piggy-back on other posts, the often get overlooked.
This one will not be answered either but for a different reason: I am not able to decode the symbols because of unknown encoding used.

I will be able to understand the symbols if your web-browser(I.E. or Firefox) had been set (using "view/encoding") to "Western (ISO-8859-1)". Alternatively, if you specify the encoding you use, I can set to the same, but my browser does not detect your encoding.