Question

Solve $2{\cos }^{2}x+4\cos x=3{\sin }^{2}x$, then $x=2n\pi \pm awherea={\cos }^{-1}\left(\frac{-2+\sqrt{19}}{m}\right),n\in I$. Find the value of $m$.

Answer 1

$\because 2{\cos }^{2}x+4\cos x=3{\sin }^{2}x=0$
$\Rightarrow 2{\cos }^{2}x+4\cos x-3(1-{\cos }^{2}x)=0$
$\Rightarrow 5{\cos }^{2}x+4\cos x-3=0$
$\Rightarrow \{\cos x-\left(\frac{-2+\sqrt{19}}{5}\right)\}\{\cos x-\left(\frac{-2-\sqrt{19}}{5}\right)\}=0$
$\because \cos x\in [-1,1]\forall x\in R\therefore \cos x\ne \frac{-2-\sqrt{19}}{5}$
$\therefore$ equation (ii) will be true if
$\cos x=\frac{-2+\sqrt{19}}{5}$
$\Rightarrow \cos x=\cos a,where\cos a=\frac{-2+\sqrt{19}}{5}$
$\Rightarrow x=2n\pi \pm awherea={\cos }^{-1}\left(\frac{-2+\sqrt{19}}{5}\right),n\in I$