16^ (Sin xCos x) = (4)^-√3Sin x 4^2Sin xCos x = 4^ - √3Sin x 2Sin xCos x = -√3Sin x 2Sin x Cos x +√3Sin x = 0 Sinx( 2Cos x + √3) = 0 а) Sin x = 0 или б) 2Cos x + √3 = 0 x = πn,где n∈Z 2Cos x = -√3 Cos x = - √3/2 x = +- arcCos(-√3/2) + 2πk,где к ∈Z x = +- 5π/6 + 2πк, где к∈Z Теперь ищем корни на отрезке [ 2π; 7π/2] a) n =1 б) k = 1 x = π x = 5π/ 6 + 2π n = 2 х = -5π/6 + 2π x = 2π k = 2 n = 3 x = 5π/6 + 4π x = 3π x = - 5π/6 + 4π = 19π/6 n = 4 k = 3 x = 4π x = 5π/6 + 6 π
4^2Sin xCos x = 4^ - √3Sin x
2Sin xCos x = -√3Sin x
2Sin x Cos x +√3Sin x = 0
Sinx( 2Cos x + √3) = 0
а) Sin x = 0 или б) 2Cos x + √3 = 0
x = πn,где n∈Z 2Cos x = -√3
Cos x = - √3/2
x = +- arcCos(-√3/2) + 2πk,где к ∈Z
x = +- 5π/6 + 2πк, где к∈Z
Теперь ищем корни на отрезке [ 2π; 7π/2]
a) n =1 б) k = 1
x = π x = 5π/ 6 + 2π
n = 2 х = -5π/6 + 2π
x = 2π k = 2
n = 3 x = 5π/6 + 4π
x = 3π x = - 5π/6 + 4π = 19π/6
n = 4 k = 3
x = 4π x = 5π/6 + 6 π