Question

1-cosa+cos2a/sina-sin2a

Answer 1

$\dfrac{1-cosa-cos2a}{sina-sin2a}=\dfrac{1-cosa-2cos^2a+1}{sina(1-cosa)}=\\\boxed{\left\{{{-2cos^2a-cosa+2=0}\atop{sina(1-cosa)\ne0}} \right. }\\1)sina(1-cosa)\ne0\\sina\ne0\ our\ cosa\ne1\\a\ne\pi k,k\in Z;a\ne \pi k,k\in Z\\ \\2)2cos^2a+cosa-2=0\\\mathbf{cos\alpha=a,a\le1}\\2a^2+a-2=0\\D=1-4*2*(-2)=17\\a_1=\dfrac{-1+\sqrt{17}}{2*2}=\dfrac{\sqrt{17}-1}{4}\\a_2= \dfrac{-1-\sqrt{17}}{2*2}=-\dfrac{\sqrt{17}+1}{4}-ne\ yd.\ ysl.\\cosa=\dfrac{\sqrt{17}-1}{4}\\a=\pm arccos(\dfrac{\sqrt{17}-1}{4})+2\pi k,k\in Z$