Bài 2
3 tan ( x - π/4 ) = căn 2
<=> tan(x - π/4) = căn 2 /3
<=> tan ( x - π/4) = arctan căn 2/3)
<=> x - π/4 = arctan ( căn 2/3)
<=> x = arctan(căn 2/3) + π/4 + k π
Bài 3
cos ( 2x + 10 độ) = - căn 3/2
<=> cos ( 2x + 10 độ) = cos ( 5 π/6)
<=> 2x + 10 độ = 5 π/6 + k2 π
hoặc 2x + 10 độ = -5 π/6 +k2 π
<=> 2x = 5 π/6 - π/18 + k2π
hoặc 2x = -5π/6 -π/18 + k2π
<=> x = 7/18 π + kπ
hoặc x = - 8π/18 + kπ

4)
4 cot (x -π/3) + 4 = 0
4 cot(x-π/3) = -4
cot (x - π/3) = -1
cot (x -π/3) = cot ( -π/4)
x -π/3 = -π/4 + kπ
x = -π/4 + π/3 + kπ
x = π/12 + kπ

7. 3sin^2(2x) + sin^2(2x) - 4 = 0
<=> 4sin^2(2x) - 4 = 0
<=> 2(1 - cos4x) - 4 = 0
<=> 2 - 2cos4x - 4 = 0
<=> cos4x = -1
<=> 4x = π + k2π, k nguyên
<=> x = π/4 + kπ/2, k nguyên
8. cos^2(5x) - sin5x - 2 = 0
<=> 1 - sin^2(5x) - sin5x - 2 = 0
<=> sin^2(5x) + sin(5x) + 1 = 0
<=> (sin5x + 1/2)^2 + 3/4 = 0, Vô lí
Vậy PT vô nghiệm

4) sin 4x + cos 4x = -1
<=> 1/ căn 2 . sim 4x + 1/căn 2 . cos 4x = - 1 (1)
đặt sin a = 1/căn 2
cos a = 1/căn 2
=> a= π/4
(1) tương đương:
sinπ/4 sin4x + cosπ/4 cos 4x = -1
<=> cos (π/4 - 4x) = -1
cos(π/4 - 4x ) = cosπ
π/4 - 4x = π +k2π
<=> - 4x = π - π/4 + k2π
-4x = 3/4 π + k2π
x = -3/16 π -kπ/2

5. sin4x + cos4x = -1
<=> sin4x/√2 + cos4x/√2 = -1/√2
<=> sin4x.cosπ/4 + cos4x.sinπ/4 = sin(-π/4)
<=> sin(4x + π/4) = sin(-π/4)
<=> 4x + π/4 = -π/4 + k2π
hoặc 4x + π/4 = 5π/4 + k2π, k nguyên
<=> 4x = -π/2 + k2π
hoặc 4x = π + k2π, k nguyên
<=> x = -π/8 + kπ/2
hoặc x = π/4 + kπ/2, k nguyên

Câu trên là bài 5
6) căn 3 cos 3x - sin 3x = căn 3
căn 3/2 cos 3x - 1/2 sin 3x = căn 3 /2 (2)
đặt cos a = căn 3/2
sina = 1/2
=> a= π/6
(2) <=> cosπ/6 .cos 3x - sin π/6 . sin3x = căn 3/2
<=> cos (π/6 + 3x ) = căn 3/2
ososs (π/6 + 3x)= cos π/6
<=> π/6 + 3x = π/6 + k2π
hoặc π/6 + 3x = -π/6 + k2π
<=> 3x = = k2π
hoặc 3x = -π/3 + k2π
<=> x = k2/3 π
hoặc x = -π/9 + k2/3π

6. √3.cos3x - sin3x = √3
<=> √3/2 . cos3x - 1/2.sin3x = √3.2
<=> sinπ/3.cos3x - cosπ/3.sin3x = sinπ/3
<=> sin(π/3 - 3x) = sinπ/3
<=> π/3 = π/3 - 3x + k2π
hoặc π/3 = 3x + 2π/3 + k2π, k nguyên
<=> 3x = k2π
hoặc 3x = -π/3 - k2π, k nguyên
<=> x = k2π/3
hoặc x = -π/9 - k2π/3, k nguyên

9)
tan^2 7x + tan 7x - 2 = 0
đặt t = tan 7x
=> t^2 + t - 2 = 0
t^2 + 2t - t - 2 = 0
t(t + 2) - (t+2) = 0
( t- 1)( t+2) = 0
<=> t- 1 = 0
t+ 2 = 0
<=> t= 1
hoặc t = -2
=>
<=> tan 7x = 1
hoặc tam 7x = - 2
<=> tan 7x = tanπ/4
hoặc tan 7x = arctan (-2)
<=> x = π/28 + kπ
hoặc x = arctan(2)/7 + kπ