Determine the zeros of the function y = k/2(sin(x)); 0 ≤ x ≤ π
y=k2(sinx);0≤x≤πy = \dfrac k2(\sin x); 0 ≤ x ≤ πy=2k(sinx);0≤x≤π
Put y=0
⇒k2(sinx)=0⇒sinx=0⇒x=0, x=π\Rightarrow \dfrac k2(\sin x)=0 \\\Rightarrow \sin x=0 \\\Rightarrow x=0,\,x=\pi⇒2k(sinx)=0⇒sinx=0⇒x=0,x=π
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