Đáp án: Vô nghiệm
Giải thích các bước giải:
Ta có:
$\dfrac{1+\cot2x\tan x}{\cos^2x}+1=6(1-\dfrac12\sin^22x)$
$\to \dfrac{1+\dfrac{\cos2x}{\sin2x}\cdot \dfrac{\sin x}{\cos x}}{\cos^2x}+1=6(1-\dfrac12\cdot (2\sin x\cos x))$
$\to \dfrac{1+\dfrac{\cos2x}{2\sin x\cos x}\cdot \dfrac{\sin x}{\cos x}}{\cos^2x}+1=6(1-2\sin^2 x\cos^2x)$
$\to \dfrac{1+\dfrac{\cos2x}{2\cos^2x}}{\cos^2x}+1=6(1-2\sin^2 x\cos^2x)$
$\to \dfrac{\dfrac{2\cos^2x+\cos2x}{2\cos^2x}}{\cos^2x}+1=6(1-2(1-\cos^2x)\cos^2x)$
$\to \dfrac{\dfrac{2\cos^2x+2\cos^2x-1}{2\cos^2x}}{\cos^2x}+1=6(1-2(1-\cos^2x)\cos^2x)$
$\to \dfrac{\dfrac{4\cos^2x-1}{2\cos^2x}}{\cos^2x}+1=6(1-2(1-\cos^2x)\cos^2x)$
$\to\dfrac{4\cos^2x-1}{2\cos^4x}+1=6(1-2(1-\cos^2x)\cos^2x)$
$\to\dfrac{2\cos^2x-1}{\cos^4x}+1=6(1-2(1-\cos^2x)\cos^2x)$
Đặt $\cos^2x=t, 0\le t\le 1$
$\to\dfrac{2t-1}{t^2}+1=6(1-2(1-t)t)$
$\to 2t-1+t^2=6\left(1-2\left(1-t\right)t\right)t^2$
$\to 2t-1+t^2=6t^2-12t^3+12t^4$
$\to 12t^4-12t^3+5t^2-2t+1=0$
$\to (12t^4-12t^3+3t^2)+t^2+(t^2-2t+1)=0$
$\to 3(4t^4-4t^3+t^2)+t^2+(t^2-2t+1)=0$
$\to 3(2t^2-t)^2+t^2+(t-1)^2=0$
Mà $3(2t^2-t)^2+t^2+(t-1)^2>0,\quad\forall 0\le t\le 1$
$\to$Phương trình vô nghiệm
