1/sinx-cosx/sinx=(1-cosx)/sinx=1/3.
cosx=1-2sin^2(x/2), so 1-cosx=1-(1-2sin^2(x/2))=2sin^2(x/2).
sinx=2sin(x/2)cos(x/2); so 2sin^2(x/2)/2sin(x/2)cos(x/2)=sin(x/2)/cos(x/2)=tan(x/2)=1/3.
The hypotenuse of the corresponding right-angled triangle is sqrt(1^2+3^2)=sqrt(10).
Therefore, sin(x/2)=1/sqrt(10); cos(x/2)=3/sqrt(10). sinx=2sin(x/2)cos(x/2)=2*3/10=3/5; cosx=4/5 (sqrt(1-(3/5)^2)).
cos(2x)=2cos^2(x)-1=2*16/25-1=(32-25)/25=7/25.