Công thức tính: √A2 + B2 =
vd: √(√3)2
+ 12 = 2
a>
√3cosx
- sinx = √2
ó√3/2 cosx – ½ sinx = 2/2
ócos3/2 cosx – sin½ sinx = 1
ócos π/2 cosx – sin π/6 sinx = 1 =>
[cos π/2 cosx – sin π/6 sinx]
ó -cos (π/6 – x) = 1
ó -cos (π/6 – x) = 0
ó π/6 – x = 0 + k2π
π/6 – x = -0 + k2π
ó
- x = - π/6 + k2π
- x = - π/6 + k2π
ó
x = π/6 + k2π
x = π/6 + k2π
(k ∈ ℤ)
b>
Cosx - √3sinx
= -1
ó ½ cosx - √3/2
sinx = -1/2
ó sin ½ cosx – cos √3/2
sinx = -1/2
ó sin π/6 cosx - cos π/6
sinx = -1/2 => [sin π/6 cosx - cos π/6 sinx]
ó sin (π/6 – x) = - π/6
ó π/6 – x = - π/6 + k2 π
π/6 – x = π + π/6 + k2 π
ó - x = - π/6 – π/6 + k2π
- x = π + π/6 –
π/6 + k2π
ó x = - π/3 + k2π
x = π +
k2π (k ∈ ℤ)
c>
sinx + √3cosx
= 2
ó ½ sinx + √3/2
cosx = 2/2
ó cos ½ sinx + sin √3/2
cosx = 1
ó cos π/3 sinx + sin π/3
cosx = 1 => [
sinx.cos π/3 + cosx.sin π/3]
ó sin (x + π/3) = 1
ó sin (x + π/3) = π/2
ó x + π/3 = π/2 + k2π
x + π/3 = π – π/2 + k2π
ó x = π/2 – π/3 + k2π
x = π – π/2 – π/3 + k2π
ó x = π/6 + k2π
x = π/6 + k2π (k ∈ ℤ)
d>
3sinx + 4cosx = 5
ó 3/5 sinx + 4/5 cosx =
5/5
ó cos 3/5 sinx + sin 4/5
cosx = 1
Bấm SHIFT
cos 3/5 và sin 4/5 sẽ ra số lẽ.
Đặt cos α
= 3/5, sin α = 4/5
ó cos α.sinx + sin α.cosx
= 1
=> [sinx.cos α + cox.sin α]
\ó sin (α + x) = 1
ó sin (α + x) = π/2
ó α
+ x = π/2
+ k2 π
α
+ x = π – π/2 + k2π
ó x = - α +
π/2 + k2π
x = - α + π/2 + k2π (k ∈ ℤ)
e>
4sin3x + cos3x = 4
ó 4/√17 sin3x + 1/√17
cos3x = 4/√17
Đặt cos α = 4/√17, sin α = 1/√17
cos α.sinx + sin α.cosx = 4/√17 => [sinx.cos
+ cosx.sin α]
ó sin (α + x)= 4/√17
ó sin (α + x) = 1.33 (loại)
Vì -1 ≤ cos
≤ 1
f>
sin2x + sin2x
= ½
ó sin2x + (1 – 2cos2x)/2
= ½
ó sin2x + ½ - ½ cos2x = ½
ó sin2x – ½ cos2x = 0
ó 2√5/5 sin2x - √5/5
cos2x = 0
Đặt cos α =
2√5/5 sin2x, sin α = √5/5 cos2x
ó cos α.sin2x + sin α.cos2x
= 0 => [sin2x.cos α + cos2x.sin α]
ó sin(α+ 2x) = 0
ó α + 2x = 0 + k2π
α + 2x = π - 0 + k2π
ó 2x = - α + k2π
2x = π - α + k2π
ó x = - α/2 + k2π/2
x = π/2 – α/2
+ k2π/2 (k ∈ ℤ)
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