1. Vì -1≤ cosx ≤ 1
nên: 0 ≤ √cosx ≤ 1
⇔0+3 ≤ √cosx + 3 ≤1+3
⇔ 3≤ 3 + √cosx ≤ 4
⇔ 3≤ y≤ 4
ymax = 4 ⇒ cosx = 1 ⇒ x = k2π ( k∈Z)
ymin = 3 ⇒ cosx = 0 ⇒ x = π/2 + kπ ( k∈ Z)
2.
a) sinx = 2/5
⇔ x= arcsin( 2/5) + k2π
x= π-arcsin(2/5) + k2π ( k∈Z)
b) sin(x+ 30° ) = sin ( 2x + 55°)
⇔ x+ 30°= 2x + 55° + k360°
x+ 30°= 180°-(2x+55°) + k360°
⇔ x -2x = 55°-30° + k360°
x+ 30 °= 180° -2x - 55° + k360°
⇔ -x = 25° + k360°
x + 2x = 180°-55°-30° + k360°
⇔ x= -25° - k360°
3x= 95° + k360° ( k ∈ Z)
c) cos(2x+1)=1
⇔ 2x+1 = k2π
⇔ 2x= -1 + k2π
⇔ x = -1/2 + kπ ( k∈Z)
d) cos(x-2)= sin3(x-π/2)
⇔ cos (x-2) = sin( 3x- 3π/2)
⇔ cos ( x-2) = cos [ π/2 -( 3x- 3π/2)]
⇔ cos (x-2) = cos ( -3x + 2π)
⇔ x-2 = -3x+2π + k2π
x-2 = - ( -3x+2π) + k2π
⇔ x + 3x = 2 + 2π + k2π
x-2 = 3x - 2π + k2π
⇔ 4x = 2 + 2π + k2π
x-3x= 2 -2π + k2π
⇔ x = 1/2 + π/2 + kπ/2
-2x= 2-2π+k2π
⇔ x = 1/2 + π/2 + kπ/2
x= -1 + π - kπ ( k ∈ Z)
