Question

A root of the equation on L.H.S. satisfied the ralation in R.H.S. Match the entries on L.H.A. and R.H.S.

Answer 1

$2(1-2si{n}^{2}x)-2sinxcos-2si{n}^{2}x=0$
$3si{n}^{2}x+sinxcos-1=0$
Diveded ny $co{s}^{2}x$
$\therefore 3ta{n}^{2}x+tanx-(1+ta{n}^{2}x)=0$
or $2ta{n}^{2}x+tanx-1=0$
or $(tanx+1)(2tanx-1)=0$
$\therefore tanx=-1,\frac{1}{2}$ $\therefore cos2x=\frac{1-t^2}{1+t^2}=0,\frac{3}{5}$
$\therefore$ (a) $\to$ (p,q,s)
(b) $2si{n}^2\left(\frac{\pi }{2}co{s}^{2}x\right)=2si{n}^2\left(\frac{\pi sin2x}{2}\right)$
$\therefore co{s}^{2}x=sin2x=2sinxcosx$
or $cosx(cosx-2sinx)=0$
$\therefore cosx=0$ or $tanx=\frac{1}{2}$
$cos2x=\frac{1-t^2}{1+t^2}=\frac{1-\left(1/4\right)}{1+\left(1/4\right)}=\frac{3}{5}$
$\therefore$ (b) $\to$ (q,s)
(c) $6(1+t^2)-11t-2=0$ or $6t^2-11t+4=0$
$therefore(2t-1)(3t-4)=0$
$\therefore t=\frac{1}{2},\frac{4}{3}⟹=$(q,r)
Also when $tanx=\frac{1}{2},cos2x=\frac{3}{5}$ as in (b)
$\therefore$ (c) $\to$ (s)
$\therefore$ (c) $\to$ (q,r,s)
(d) Divided by $co{s}^{2}x$
$\therefore 3se{c}^{2}x-tanx-7=0$
$3(1+t^2)-t-7=0$ or $3t^2-t-4=0$
or $(3t-4)(t+1)=0$ $\therefore t=tanx=-1,\frac{4}{3}$
$\therefore$ (d) $\to$ (p,r)