Question

Solve the following equation:

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

Answer 1

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

$$\Rightarrow 4-4{\cos }^{2}x-8\cos x+1=0$$
$$\Rightarrow 4{\cos }^{2}x+8\cos x-5=0$$
$$\Rightarrow 4{\cos }^{2}x+10\cos x-2\cos x-5=0$$
$$\Rightarrow 2\cos x(2\cos x+5)-1(2\cos x+5)=0$$
$$\Rightarrow (2\cos x-1)(2\cos x+5)=0$$

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

$$(2\cos x+5)=0$$

Now,

$$2\cos x+5=0\Rightarrow \cos x=-\frac{5}{2}$$  (It is not possible.)

$$\therefore 2\cos x-1=0$$

$$\Rightarrow \cos x=\frac{1}{2}$$

$$\Rightarrow \cos x=\cos \frac{\pi }{3}$$

$$\Rightarrow x=2n\pi \pm \frac{\pi }{3},n\in Z$$