Question

A real number $\theta$ is a root of the equation $7co{s}^{2}x+4cosx-1=0$. If $pco{s}^2\theta +qco{s}^2\theta +r=0$ where p,q,r are real constants then
r can be

Answer 1

$7co{s}^2\theta +4cos\theta -1=0cos\theta =\frac{-4\pm \sqrt{4^2-4\left(7\right)(-1)}}{2\left(7\right)}=\frac{-4\pm \sqrt{16+28}}{14}=\frac{-4\pm \sqrt{44}}{14}=\frac{-4\pm 2\sqrt{11}}{14}=\frac{-2\pm 2\sqrt{11}}{7}$
$cos2\theta =2co{s}^2\theta -1cos2\theta =2{\left(\frac{-2\pm \sqrt{11}}{7}\right)}^2-1cos2\theta =\frac{2}{49}(-2\pm \sqrt{11}{)}^2-1=\frac{2}{49}(4+11\pm 4\sqrt{11})-1=\frac{2(15\pm 4\sqrt{11})-49}{49}=\frac{60\pm 8\sqrt{11}-49}{49}=\frac{-19\pm 8\sqrt{11}}{49}co{s}^{2}2\theta ={\left(\frac{-19\pm 8\sqrt{11}}{49}\right)}^2$
$pco{s}^{2}2\theta +qco{s}^{2}2\theta +r=0p{\left(\frac{-19\pm 8\sqrt{11}}{49}\right)}^2+q\left(\frac{-19\pm 8\sqrt{11}}{49}\right)+r=0\frac{-19\pm 8\sqrt{11}}{49}=\frac{-q\pm \sqrt{q^2-4\left(p\right)\left(r\right)}}{2p}$
By comparing we get $2p=49$
$p=\frac{49}{2}...\left(i\right)$
$-19=-q\to q=19\to \left(ii\right)\pm \sqrt{64811}=\pm \sqrt{q^2-4pr}\pm \sqrt{704}=\pm \sqrt{q^2-4pr}\Rightarrow 704=q^2-4pr704={19}^2-4\left(\frac{49}{2}\right)r704=361-98r98r=361-704r=\frac{-343}{98}r=-3.5...\left(iii\right)$
$\Rightarrow pco{s}^{2}2\theta +qcos2\theta +r=0\frac{49}{2}co{s}^{2}2\theta +19cos2\theta -3.5=0\to \left(iv\right)$
Multiplying (iv) by (ii)
$\Rightarrow 49co{s}^{2}2\theta +38cos2\theta -7=0\to \left(v\right)From\left(iii\right)p=49q=38r=-7$.