Câu 1:
sinx = 1/2
<=> x = π/6 + k2π hay x = π - π/6 + k2π
<=> x = π/6 + k2π hay x = 5π/6 + k2π
Vậy...
Câu 2:
sin2x + √3cos2x = 2
<=> 1/2.sin2x + √3/2.cos2x = 1
<=> sin(2x + π/3) = 1
<=> 2x + π/3 = π/2 + k2π
<=> 2x = π/6 + k2π
<=> x = π/12 + k2π
Vậy....

Câu 3:
(sin2x + cos2x)cosx + 2cos2x - sinx = 0
<=> sin2xcosx + cos2xcosx + 2cos2x - sinx = 0
<=> 2sinxcos^2(x) - sinx + cos2xcosx + 2cos2x = 0
<=> sinx(2cos^2(x) - 1) + cos2x(cosx + 2) = 0
<=> sinxcos2x + cos2x(cosx + 2) = 0
<=> cos2x(sinx + cosx + 2) = 0
<=> cos2x = 0 hay sinx + cosx = - 2
<=> 2x = π/2 + kπ hay sin(x + π/4) = - √2 (VN)
<=> x = π/4 + kπ/2
Vậy...

Câu 1:
sinx = 1/2
<=> sinx = sin π/6
<=> x = π/6 + k2π
hoặcx = π - π/6 + k2π
<=> x = π/6 + k2π
hoặc x = 5π/6 + k2π
Vậy...
Câu 2:
sin2x + √3cos2x = 2
<=> 1/2.sin2x + √3/2.cos2x = 1
<=> sin(2x + π/3) = 1
<=> 2x + π/3 = π/2 + k2π
<=> 2x = π/6 + k2π
<=> x = π/12 + k2π
Vậy....
Câu 3:
(sin2x + cos2x)cosx + 2cos2x - sinx = 0
<=> sin2xcosx + cos2xcosx + 2cos2x - sinx = 0
<=> 2sinxcos^2(x) - sinx + cos2xcosx + 2cos2x = 0
<=> sinx(2cos^2(x) - 1) + cos2x(cosx + 2) = 0
<=> sinxcos2x + cos2x(cosx + 2) = 0
<=> cos2x(sinx + cosx + 2) = 0
<=> cos2x = 0
hoặc sinx + cosx = - 2
<=> 2x = π/2 + kπ
hoặc sin(x + π/4) = - √2 ( loại)
<=> x = π/4 + kπ/2
Vậy...