`\sqrt{3}. sinx - cosx = \sqrt{3}. cos2x - sin2x`  `(D=RR)`

`⇔ \sqrt{3}/2 sinx - 1/2 cos x = \sqrt{3}/2 cos2x - 1/2 sin2x`

`⇔ sin(x- pi/6) = sin(pi/3 -2x)`

$⇔\left[\begin{matrix} x-\dfrac{\pi}{6}=\dfrac{\pi}{3} -2x+k2\pi\\ x-\dfrac{\pi}{6}=\pi-\dfrac{\pi}{3}+2x+k2\pi\end{matrix}\right.$

$⇔\left[\begin{matrix} x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}(k∈Z)(nhận)\\ x=\dfrac{-5\pi}{6} + k2\pi(k∈Z)(nhận)\end{matrix}\right.$

`**) x=\frac{\pi}{6}+\frac{k2\pi}{3}`
`<=> (-3pi)/2 ≤\frac{\pi}{6}+\frac{k2\pi}{3}≤2pi`
`<=> -5/2 ≤ k ≤ 11/4`
Mà `k∈ZZ`
`-> k∈{-2;-1;0;1;2}`

`**) x=\frac{-5\pi}{6} + k2\pi`
`<=> (-3pi)/2 ≤\frac{-5\pi}{6} + k2\pi≤2pi`
`<=> -1/3 ≤ k ≤ 17/12`
Mà `k∈ZZ`
`-> k∈{0;1;}`

Vậy pt có tập no `S={(-7pi)/6; (-pi)/2;pi/6;(5pi)/6;(3pi)/2;(-5pi)/6;(7pi)/6}`

Đáp án:

`S={pi/6;(5pi)/6;-pi/2;(3pi)/2;-(7pi)/6;-(5pi)/6;(7pi)/6}`

Giải thích các bước giải:

 `\sqrt3 sinx-cosx=\sqrt3 cos2x-sin2x`

`<=>(\sqrt3)/2 sinx-1/2 cosx=(\sqrt3)/2cos2x -1/2 sin2x`

`<=>sin(x-pi/6)=sin(pi/3-2x)`

`@: th1:x-pi/6=pi/3-2x+k2pi`

           `<=>x=pi/6+k(2pi)/3 (k∈ZZ)`

Vì `x∈((-3pi)/2;2pi)`

`(-3pi)/2<\frac{\pi}{6}+\frac{k2\pi}{3}<2pi`

`=>-3/2<1/6+k .2/3 <2`

`=>-5/2<k<11/4`

`=>k∈{0;±1;±2}`

`=>x∈{pi/6;(5pi)/6;-pi/2;(3pi)/2;-(7pi)/6}`

`@: th2: x-pi/6=pi-pi/3+2x+k2pi`

`<=>x=(-5pi)/6+k2pi (k∈ZZ)`

`x∈((-3pi)/2;2pi)`

`=>-3pi/2<(-5pi)/6+k2pi<2pi`

`=>-1/3 <k<17/12`

`=>k∈{0;1}`

`=>x∈{-(5pi)/6+(7pi)/6}`

 Vậy `S={pi/6;(5pi)/6;-pi/2;(3pi)/2;-(7pi)/6;-(5pi)/6;(7pi)/6}`