Đáp án:
Giải thích các bước giải:
$\dfrac{1}{4\cos ^2\left(x\right)}+2\cos \left(\dfrac{x}{2}\right)=\dfrac{1}{2\cos \left(x\right)}+\sqrt{4\cos \left(x\right)+1}$
$\dfrac{1}{4\cos ^2\left(x\right)}+2\cos \left(\frac{x}{2}\right)-\dfrac{1}{2\cos \left(x\right)}-\sqrt{4\cos \left(x\right)+1}=0$
$\dfrac{1+8\cos ^2\left(x\right)\cos \left(\dfrac{x}{2}\right)-2\cos \left(x\right)-4\cos ^2\left(x\right)\sqrt{4\cos \left(x\right)+1}}{4\cos ^2\left(x\right)}=0$
$1+8\cos ^2\left(x\right)\cos \left(\dfrac{x}{2}\right)-2\cos \left(x\right)-4\cos ^2\left(x\right)\sqrt{4\cos \left(x\right)+1}=0$
Đặt `u=x/2`, ta có:
$1+8\cos ^2\left(2u\right)\cos \left(u\right)-2\cos \left(2u\right)-4\cos ^2\left(2u\right)\sqrt{4\cos \left(2u\right)+1}=0$
Nếu `x/2 = arccos ( x/2 ) + 2pin`
`=>x=pi/3 + 4pi n`
Nếu `x/2 = 2pi - arccos (x/2) +2pin`
`=>x=4pi-pi/3 +4pin`
