<span>f(x)=ln2x найти f'(x), f'(1/2) Помогите решить пожалуйста</span>
1) -3x-5= - 10x
-3х+10х=5
7х=5
х=5\7
2) -x+10= - 7x
-х+7х=-10
6х=-10
х=-10\6
х=-5\3
х=-1 2\3
3) -x+7= - 7x
-х+7х=-7
6х=-7
х=-7\6
4) -2x+6= - 9x
-2х+9х=-6
7х=-6
х=-6\7
5) -10x-8= - 7x
-10х+7х=8
-3х=8
х=-8\3
х=-2 2\3
<span>6) -6x-8= - 3x
-6х+3х=8
-3х=8
х=-8\3
х=-2 2\3</span>
Решение
2sinxcos3x + sin4x = 0
2*(1/2)*[sin(x - 3x) + sin(x + 3x)] + sin4x = 0
- sin2x + 2sin4x = 0
2sin2x*cos2x - sin2x = 0
sin2x*(2cos2x - 1) = 0
1) sin2x = 0
2x = πk, k ∈ Z
x₁ = πk/2, k ∈ Z
2) 2cos2x - 1 = 0
cos2x = 1/2
2x = (+ -)*arccos(1/2) + 2πn, n ∈ Z
2x = (+ -)*(π/3) + 2πn, n ∈ Z
x₂ = (+ -)*(π/6 + πn, n ∈ Z