Question

Find the general solution of $4{\sin }^{2}x-8\cos x+1=0$.

Answer 1

$4{\sin }^{2}x-8\cos x+1=0$

$\Rightarrow 4(1-{\cos }^{2}x)-8\cos x+1$

$\Rightarrow 4-4{\cos }^{2}x-8\cos x+1=0$

$\Rightarrow -4{\cos }^{2}x-8\cos x+5=0$

$\Rightarrow 4{\cos }^{2}x+8\cos x-5=0$

$\Rightarrow 4{\cos }^{2}x-2\cos x+10\cos x-5=0$

$\Rightarrow 2\cos x(2\cos x-1)+5(2\cos x-1)=0$

$\Rightarrow (2\cos x-1)(2\cos x+5)=0$

$\Rightarrow \cos x=\frac{1}{2},\cos x=-\frac{5}{2}$

$\cos x=\frac{1}{2}=\cos \frac{\pi }{3}$

$\therefore x=2n\pi \pm \frac{\pi }{3}$