bài 3
a. 4sinx.cosx.cos2x = 1
<=> 2.2sinxcosx.cos2x = 1
<=> 2sin2x.cos2x = 1
<=> sin4x = 1
<=> 4x = π/2 + k2π
<=> x = π/8 + kπ/2 , k thuộc Z
vậy .....

2.
a.
sinx = cos(x + π)
<=> cos(x+π) = cos(π/2 - x)
<=> x + π = π/2 - x + k2π hoặc x + π = x - π/2 + k2π, k nguyên
<=> 2x = -π/2 + k2π hoặc 0 = -3π/2 + k2π (loại)
<=> x = -π/4 + kπ, k nguyên
b.
cos(2x - π/4) + sin(2x + π/3) = 0
<=> cos(2x - π/4) = -sin(2x + π/3
<=> cos(2x - π/4) = sin(-2x - π/3)
<=> cos(2x - π/4) = cos(5π/6 + 2x)
<=> 2x - π/4 = 5π/6 + 2x + k2π (loại) hoặc 2x - π/4 = -5π/6 - 2x + k2π ( k nguyên )
<=> 4x = -7π/12 + k2π
<=> x = -7π/48 + kπ/2, k nguyên

3.
a.
4sinx.cosx.cos2x = 1
<=> 2sin2xcos2x = 1
<=> sin4x = 1
<=> 4x = π/2 + k2π, k nguyên
<=> x = π/8 + kπ/2, k nguyên
b.
sin^2(2x) + cos^2(3x) = 1
<=> sin^2(2x) + cos^2(3x) = sin^2(2x) + cos^2(2x)
<=> cos^2(3x) = cos^2(2x)
<=> | cos3x | = | cos2x |
<=> cos3x = cos2x hoặc cos3x = -cos2x
<=> cos3x = cos2x hoặc cos3x = cos(π - 2x)
<=> 3x = 2x + k2π hoặc 3x = -2x + k2π
hay 3x = π - 2x + k2π hoặc 3x = 2x - π + k2π
<=> x = k2π hoặc x = k2π/5 hoặc x = π/5 + k2π/5 hoặc x = -π + k2π, k nguyên

3.
c.
sin^2(5x + 2π/5) = cos^2(x/4 + π)
<=> cos^2(π/10 - 5x) = cos^2(x/4 + π)
<=> | cos(π/10 - 5x) | = | cos(x/4 + π |
<=> cos(π/10 - 5x) = cos(x/4 + π) hoặc cos(π/10 - 5x) = -cos(x/4 + π) = cos(-x/4) = cos(x/4)
<=> x/4 + π = π/10 - 5x + k2π hoặc x/4 + π = 5x - π/10 + k2π
hoặc x/4 = π/10 - 5x + k2π hoặc x/4 = 5x - π/10 + k2π
<=> x = -6π/35 + 8π/21 hoặc x = 22π/95 - 8π/19
hoặc x = 2π/105 + 8π/21 hoặc x = 2π/95 - 8π/19; k nguyên

3.
e.
sin^6(x) + cos^6(x) = 0
<=> (sin^2x + cos^2x)^3 - 3sin^2x.cos^2x(sin^2x + cos^2x) = 0
<=> 1^3 - 3sin^2x.cos^2x.1 = 0
<=> 1 - 3(sinx.cosx)^2 = 0
<=> 1 - 3/4.sin^2(2x) = 0
<=> 3/8.(1-2sin^2(2x)) + 5/8 = 0
<=> 3/8.cos4x + 5/8 = 0
<=> cos4x = -5/3, PT VN do cos4x > -1 với mọi x
Vậy PT đã cho vô nghiệm