`{1+sin^2 x}/{1-sin^2 x} -tan^2 x =4 (1)`
Đk: `cosx≠0<=>x≠π/2 +kπ(k∈Z)`
`(1)<=>{1+sin^2 x}/{cos^2 x} -{sin^2 x}/{cos^2 x} =4`
`<=>{1+sin^2 x-sin^2 x}/{cos^2 x}=4`
`<=>1/{cos^2 x}=4`
`<=>tan^2 x+1=4`
`<=>tan^2 x=3`
$⇔\left[\begin{array}{l}tanx=\sqrt{3}\\tanx=-\sqrt{3}\end{array}\right.$
$⇔\left[\begin{array}{l}x=\dfrac{π}{3} +kπ\\x=-\dfrac{π}{3}+kπ\end{array}\right.$ `(k∈Z)`
`x∈[0;2π]`
$⇒\left[\begin{array}{l}0≤\dfrac{π}{3}+kπ≤2π\\0≤-\dfrac{π}{3}+kπ≤2π\end{array}\right.$ `(k∈Z)`
$⇔\left[\begin{array}{l}\dfrac{-1}{3}≤k≤\dfrac{5}{3}\\\dfrac{1}{3}≤k≤\dfrac{7}{3}\end{array}\right.$ `(k∈Z)`
$⇔\left[\begin{array}{l}k=0;k=1⇒x=\dfrac{π}{3};x=\dfrac{4π}{3}\\k=1;k=2⇒x=\dfrac{2π}{3};x=\dfrac{5π}{3}\end{array}\right.$
Ta có:
`π/3 + {4π}/3 + {2π}/3 + {5π}/3 =4π`
`text{Vậy tổng các nghiệm trong đoạn [0;2π] là 4π}`
