Apr 28, 2012 #2 C CaptainBlack Well-known member Jan 26, 2012 890 Re: quad equation with trigo jacks said: Click to expand... Do you mean has more than two roots as an equation in \(x\)? Well if this is a non-degenerate quadratic it has exactly two roots in \(\mathbb{C}\), and two or fewer roots in \( \mathbb{R}\). For it to have more than two roots all of the coefficients (including the constant term) must be zero. CB
Re: quad equation with trigo jacks said: Click to expand... Do you mean has more than two roots as an equation in \(x\)? Well if this is a non-degenerate quadratic it has exactly two roots in \(\mathbb{C}\), and two or fewer roots in \( \mathbb{R}\). For it to have more than two roots all of the coefficients (including the constant term) must be zero. CB
Apr 28, 2012 Thread starter #3 J jacks Well-known member Apr 5, 2012 226 Re: quad equation with trigo Using Caption Black Hint If $Ax^2+Bx+C=0$ has more then Two Roots, Then It will become an Identity Which is True for all Real $x$ So $A=B=C=0$ Now here $\left(a-\sin \theta\right)\alpha^2+b\alpha+\left(c+\cos \theta\right) = 0$ Similarly $\left(a-\sin \theta\right)\beta^2+b\beta+\left(c+\cos \theta\right) = 0$ and $\left(a-\sin \theta\right)\gamma^2+b\gamma+\left(c+\cos \theta\right) = 0$ Now We Can in General as $\left(a-\sin \theta\right)y^2+by+\left(c+\cos \theta\right) = 0$ Where $y=\alpha\;,\beta\;,\gamma$ are the roots of above Given equation If This equation has Real and Distinct Roots, then Its Discriminant $>0$ So $b^2-4.\left(a-\sin \theta\right).\left(c+\cos \theta\right)>0$ and $a+b+c=1$ is Given after Simplification $(b^2-4ac)+4(c\sin \theta+a\cos \theta-\sin \theta.\cos \theta)>0$ Now How can find value of $\theta$ from here Help Required. Thanks
Re: quad equation with trigo Using Caption Black Hint If $Ax^2+Bx+C=0$ has more then Two Roots, Then It will become an Identity Which is True for all Real $x$ So $A=B=C=0$ Now here $\left(a-\sin \theta\right)\alpha^2+b\alpha+\left(c+\cos \theta\right) = 0$ Similarly $\left(a-\sin \theta\right)\beta^2+b\beta+\left(c+\cos \theta\right) = 0$ and $\left(a-\sin \theta\right)\gamma^2+b\gamma+\left(c+\cos \theta\right) = 0$ Now We Can in General as $\left(a-\sin \theta\right)y^2+by+\left(c+\cos \theta\right) = 0$ Where $y=\alpha\;,\beta\;,\gamma$ are the roots of above Given equation If This equation has Real and Distinct Roots, then Its Discriminant $>0$ So $b^2-4.\left(a-\sin \theta\right).\left(c+\cos \theta\right)>0$ and $a+b+c=1$ is Given after Simplification $(b^2-4ac)+4(c\sin \theta+a\cos \theta-\sin \theta.\cos \theta)>0$ Now How can find value of $\theta$ from here Help Required. Thanks