quadratic equation with trigo

[jacks]

 

[CaptainBlack]

Re: quad equation with trigo

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

 

[jacks]

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