Đáp án + Giải thích các bước giải:
`d.`
`tan^4x-4tan^2x+3=0`
`<=>tan^4x-3tan^2x-tan^2x+3=0`
`<=>tan^2x(tan^2x-3)-(tan^2x-3)=0`
`<=>(tan^2x-1)(tan^2x-3)=0`
`<=>`\(\left[ \begin{array}{l}\tan^2x=1\\\tan^2x=3\end{array} \right.\)`<=>`\(\left[ \begin{array}{l}\tan x=±1\\\tan x=±\sqrt3\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=±\dfrac{\pi}{4}+k\pi\\x=±\dfrac{\pi}{3}+k\pi\end{array} \right.\) `(kinZZ)`
`f.`
`sin^4x+cos^4x=sin2x-1/2`
`<=> (sin^2x + cos^2x)^2 - 2sin^2xcos^2x = sin2x - 1/2 `
`<=> 1 - 1/2(2sinx.cosx)^2 = sin2x - 1/2 `
`<=> 2 - (2sinx.cosx)^2 = 2sin2x - 1 `
`<=>2-sin^2 2x=2sin2x-1`
`<=> sin^2 2x + 2sin2x - 3 = 0 `
`<=> sin^2 2x -sin2x +3sin2x- 3 = 0 `
`<=>sin2x(sin2x-1)+3(sin2x-1)=0`
`<=>(sin2x-1)(sin2x+3)=0`
`<=>`\(\left[ \begin{array}{l}\sin2x=1\\\sin2x=-3(VN)\end{array} \right.\)
`<=>2x=pi/2+k2pi`
`<=>x=pi/4+kpi(kinZZ)`