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
6x² + 6/x² + 5x + 5/x - 38 = 0
6(x² + 1/x²) + 5(1/x + x) - 38 = 0
x ≠ 0
замена
1/x + x = t
(1/x + x)² = t²
1/x² + 2*1/x * x + x² = t²
1/x² + 2 + x² = t²
1/x² + x² = t² - 2
6(x² + 1/x²) + 5(1/x + x) - 38 = 0
6(t² - 2) + 5t - 38 = 0
6t² - 12 + 5t - 38 = 0
6t² + 5t - 50 = 0
D = 25 + 4*50*6 = 1225 = 35²
t12 = (-5 +- 35)/12 = 30/12 (5/2) - 40/12 (-10/3)
обратно к х
1. 1/x + x = 5/2
2x² - 5x + 2 = 0
D = 25 - 16 = 9 = 3²
x12 = (5 +- 3)/4 = 2 1/2
2. 1/x + x = -10/3
3x² + 10x + 3 = 0
D = 100 - 36 = 64 = 8²
x12 = (-10 +- 8)/6 = -3 -1/3
ответ x = {2,1/2,-3,-1/3}
вкратце