`a)` `sinx+cosx-4sinx.cosx-1=0` $(1)$
Đặt `a=sinx+cosx` `(|a|\le \sqrt{2})`
`=>a^2=sin^2x+cos^2x+2sinxcosx`
`=>a^2=1+2sinxcosx`
`=>2sinxcosx=a^2-1`
$\\$
`(1)<=>a-2.(a^2-1)-1=0`
`<=>a-2a^2+2-1=0`
`<=>-2a^2+a+1=0`
`<=>`$\left[\begin{array}{l}a=1\\a=\dfrac{-1}{2}\end{array}\right. \ (thỏa\ đk)$
+) `TH1: a=1`
`<=>sinx+cosx=1`
`<=>\sqrt{2}/2 sinx +\sqrt{2}/2 cosx=\sqrt{2}/2`
`<=>cos\ π/4 sinx+sin\ π/4 cosx=sin\ π/4`
`<=>sin(x+π/4)=sin\ π/4`
`<=>`$\left[\begin{array}{l}x+\dfrac{π}{4}=\dfrac{π}{4}+k2π\\x+\dfrac{π}{4}=π-\dfrac{π}{4}+k2π\end{array}\right. (k\in Z)$
`<=>`$\left[\begin{array}{l}x=k2π\\x=\dfrac{π}{2}+k2π\end{array}\right. \ (k\in Z)$
$\\$
+) `TH2: a=-1/ 2`
`<=>sinx+cosx=-1/2`
`<=>\sqrt{2}/2 sinx +\sqrt{2}/2 cosx={-\sqrt{2}}/4`
`<=>cos\ π/4 sinx+sin\ π/4 cosx={-\sqrt{2}}/4`
`<=>sin(x+π/4)={-\sqrt{2}}/4`
`<=>`$\left[\begin{array}{l}x+\dfrac{π}{4}=arcsin\dfrac{-\sqrt{2}}{4}+k2π\\x+\dfrac{π}{4}=π-arcsin\dfrac{-\sqrt{2}}{4}+k2π\end{array}\right. (k\in Z)$
`<=>`$\left[\begin{array}{l}x=\dfrac{-π}{4}+arcsin\dfrac{-\sqrt{2}}{4}+k2π\\x=\dfrac{3π}{4}-arcsin\dfrac{-\sqrt{2}}{4}+k2π\end{array}\right. (k\in Z)$
Vậy: $\left[\begin{array}{l}x=k2π\\x=\dfrac{π}{2}+k2π\\x=\dfrac{-π}{4}+arcsin\dfrac{-\sqrt{2}}{4}+k2π\\x=\dfrac{3π}{4}-arcsin\dfrac{-\sqrt{2}}{4}+k2π\end{array}\right. (k\in Z)$
$\\$
`b)` `sin^3x-\sqrt{3} cos^3x=sinx.cos^2x-\sqrt{3} sin^2x.cosx` $(2)$
Nếu ` cosx=0=>x=π/2+kπ\ (k\in ZZ)`
`=>sinx=±1`
+) `sinx=1`
`(2)<=>1=0` (vô lý)
+) `sinx=-1`
`(2)<=>-1=0` (vô lý)
`=>x\ne π/2+kπ\ (k\in ZZ)`
`=>cosx\ne 0=>cos^3x\ne 0`
$\\$
Chia hai vế của `(2)` cho `cos^3 x`
`(2)<=>({sinx}/{cosx})^3-\sqrt{3}={sinx}/{cosx}-\sqrt{3} ({sinx}/{cosx})^2`
`<=>tan^3x-\sqrt{3}=tanx-\sqrt{3}tan^2x`
`<=>tan^3x+\sqrt{3}tan^2x-tanx-\sqrt{3}=0`
`<=>tan^2x(tanx+\sqrt{3})-(tanx+\sqrt{3})=0`
`<=>(tanx+\sqrt{3})(tan^2x-1)=0`
`<=>`$\left[\begin{array}{l}tanx=-\sqrt{3}\\tanx=1\\tanx=-1\end{array}\right.$
`<=>`$\left[\begin{array}{l}x=\dfrac{-π}{3}+kπ\\x=\dfrac{π}{4}+kπ\\x=\dfrac{-π}{4}+kπ\end{array}\right.\ (k\in Z) (thỏa\ mãn)$
