Giải PT a, 4sin3xcosx = 4sinxcosx - căn2 b, (2căn3)sinxcosx + cos2x = 0

Giải thích các bước giải+Đáp án:

 a)

`4sin3xcosx=4sinxcosx-\sqrt2`

`<=>4sin3xcosx-4sinxcosx=-\sqrt2`

`<=>4cosx(sin3x-sinx)=-\sqrt2`

`<=>4cosx.2cos2xsinx=-\sqrt2`

`<=>8sinxcosx.cos2x=-\sqrt2`

`<=>4sin2xcos2x=-\sqrt2`

`<=>2sin4x=-\sqrt2`

`<=>sin4x=-\sqrt2/2=sin(-\pi/4)`

`<=>`\(\left[ \begin{array}{l}4x=-\dfrac{\pi}{4}+k2\pi\\4x=\pi+\dfrac{\pi}{4}+k2\pi\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-\dfrac{\pi}{16}+\dfrac{k\pi}{2}\\x=\dfrac{5\pi}{16}+\dfrac{k\pi}{2}\end{array} \right.\)  `(k∈Z)`

b) `2\sqrt3sinxcosx+cos2x=0`

`<=>\sqrt3sin2x+cos2x=0`

`<=>\sqrt3/(2)sin2x+1/2cos2x=0`

`<=>cos(\pi/6)sin2x+sin(\pi/6)cos2x=0`

`<=>sin(2x+\pi/6)=0`

`<=>2x+\pi/6=k\pi`

`<=>2x=-\pi/6+k\pi`

`<=>x=-\pi/12+(k\pi)/2`  `(k∈Z)`