PT Luong Giac 11 Nang Cao !! Đẳng Cấp !!?

1)

cos⁴x + sin⁴x + cos(x - π/4)*sin(3x - π/4) - 3/2 = 0

<=>(sin²x + cos²x)(sin²x + cos²x - sinx*cosx) + 1/2[sin(4x - π/2) + sin2x] = 3/2

<=>1 - 1/2sin2x + 1/2(sin4x*cosπ/2 - cos4x*sinπ/2) + 1/2sin2x = 3/2

<=> -1/2cos4x -1/2 = 0

<=>cos4x = -1

<=>4x = π + k2π

<=>x = π/4 + kπ/2...(k∈Z)

2)

cotx + sinx*[1 + tanx*tan(x/2)] = 4

Gọi y = x/2 => x = 2y

=>cot2y + sin2y(1 + tan2y*tany) = 4

<=>cot2y + sin2y(1 + sin2y*siny/cos2y*cosy) = 4

<=>cot2y + sin2y(cos2y*cosy + sin2y*siny/cos2y*cosy) = 4

<=>cot2y + sin2y*cosy/cos2y*cosy = 4

<=>1/tan2y + tan2y = 4

<=>tan²2y - 4tan2y + 1 = 0

hay

tan²x - 4tanx + 1 = 0

=>ra kết quả theo tanx.

3)

sin²3x - cos²4x = sin²5x - cos²6x

<=>(1 - cos6x)/2 - (1 + cos8x)/2 = (1 - cos10x)/2 - (1+cos12x)/2

<=>cos6x + cos8x = cos10x + cos12x

<=>2cos7x*cosx - 2cos11x*cosx = 0

<=>2cosx(cos7x - cos11x) = 0

<=>{cosx = 0

.......{cos7x = cos11x

+cosx = 0 <=> x = π/2 + kπ..(k∈Z)

+cos7x = cos11x

<=>{11x = 7x + k2π

.......{11x = -7x + k2π

<=>{4x = k2π

.......{18x = k2π

<=>{x = kπ/2

.......{x = kπ/9....(k∈Z)

Mình chỉ biết làm câu thứ 3 thôi.

<=>1-cos6x/2 - 1+cos8x/2 = 1-cos10x/2 - 1+cos12x/2

<=>1-cos6x-1-cos8x-1+cos10x+1+cos12x = 0

<=>(cos12x - cos 6x) + (cos10x - cos8x) = 0

<=>-2sin9x.sin3x - 2sin3x.sinx = 0

<=>-2sin3x(sin3x + sinx) = 0

<=>-2sin9x.2sin2x.cosx=0

<=>3th:

sin9x=0 <=>9x=kpi <=> x = kpi/9

sin2x=0 <=> x =kpi/2

cosx=0 <=> x = pi/2 + kpi

cos^4x+ sin^4x = 1 - 2sin^2x.cos^2x

cos(x-pi/4).sin(3x-pi/4) = 1/2( sin2x + sin( -pi/2 +4x)) = 1/2( sin2x -cos4x)

vay pt la 1- 2sin^2x.cos^2x + 1/2( sin2x - cos4x) = 3/2

=> 2 - 1-cos^2(2x) + sin2x - cos4x = 3

=> 2+( 1 + cos4x ) /2 = sin2x - cos4x

=> 5 + cos4x - 2cos4x - 2sin2x =0

=> - ( 1- sin^2(2x) ) - 2sin2x + 5 =0

=> sin^2(2x) - sin2x +6 =0

3/

sin^2(3x) - cos^2(4x) = sin2(5x) - cos^2(6x)

=> 1 - côs6x - ( 1+ cos8x) = 1 - cos10x - ( 1+ cos12x)

=> côs6x + cos8x = cos10x + cos12

=> côs6x - cos12x = cos12x - cos 8x

=> -2sin9x.sin3x = -2sin9x.sin2x

=> sin9x =0 hay sin3x= sin2x

bai 2 ne

cotx + sinx( 1 + tanx.tan(x/2)) =4

dat y = x/2 => 2y = x

=> cot2y + sin2y ( cos2y.cosy + sin2y.siny)/cos2y.cosy =4

=> cot2y + sin2y.cosy/ cos2y.cosy =4

=> cot2y + tan2y =4 => 1/tan2y + tan2y =4

=> tan^2(2y) - 4tan2y + 1 =0