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Phổ biến Lượng giác >

cos^4(x)=cos^{23}(x)

Lời Giải

cos^{4} (x) = cos^{23} (x)

Lời Giải

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2 πn

Độ

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

Các bước giải pháp

cos^{4} (x) = cos^{23} (x)

Giải quyết bằng cách thay thế

cos^{4} (x) = cos^{23} (x)

Cho:

cos (x) = u

u^{4} = u^{23}

u^{4} = u^{23} : u = 0, u = 1

u^{4} = u^{23}

Đổi bên

u^{23} = u^{4}

Di chuyển

u^{4}

sang bên trái

u^{23} = u^{4}

Trừ

u^{4}

cho cả hai bên

u^{23} - u^{4} = u^{4} - u^{4}

Rút gọn

u^{23} - u^{4} = 0

u^{23} - u^{4} = 0

Hệ số

u^{23} - u^{4} : u^{4} (u - 1) (u^{18} + u^{17} + u^{16} + u^{15} + u^{14} + u^{13} + u^{12} + u^{11} + u^{10} + u^{9} + u^{8} + u^{7} + u^{6} + u^{5} + u^{4} + u^{3} + u^{2} + u + 1)

u^{23} - u^{4}

Đưa số hạng chung ra ngoài ngoặc

u^{4} : u^{4} (u^{19} - 1)

u^{23} - u^{4}

Áp dụng quy tắc số mũ:

a^{b + c} = a^{b} a^{c}

u^{23} = u^{19} u^{4}

= u^{19} u^{4} - u^{4}

Đưa số hạng chung ra ngoài ngoặc

u^{4}

= u^{4} (u^{19} - 1)

= u^{4} (u^{19} - 1)

Hệ số

u^{19} - 1 : (u - 1) (u^{18} + u^{17} + u^{16} + u^{15} + u^{14} + u^{13} + u^{12} + u^{11} + u^{10} + u^{9} + u^{8} + u^{7} + u^{6} + u^{5} + u^{4} + u^{3} + u^{2} + u + 1)

u^{19} - 1

Viết lại

1

dưới dạng

1^{19}

= u^{19} - 1^{19}

Áp dụng quy tắc phân tích thành nhân tử:

x^{n} - y^{n} = (x - y) (x^{n - 1} + x^{n - 2} y + \dots + x y^{n - 2} y^{n - 1})

u^{19} - 1^{19} = (u - 1) (u^{18} + u^{17} + u^{16} + u^{15} + u^{14} + u^{13} + u^{12} + u^{11} + u^{10} + u^{9} + u^{8} + u^{7} + u^{6} + u^{5} + u^{4} + u^{3} + u^{2} + u + 1)

= (u - 1) (u^{18} + u^{17} + u^{16} + u^{15} + u^{14} + u^{13} + u^{12} + u^{11} + u^{10} + u^{9} + u^{8} + u^{7} + u^{6} + u^{5} + u^{4} + u^{3} + u^{2} + u + 1)

= u^{4} (u - 1) (u^{18} + u^{17} + u^{16} + u^{15} + u^{14} + u^{13} + u^{12} + u^{11} + u^{10} + u^{9} + u^{8} + u^{7} + u^{6} + u^{5} + u^{4} + u^{3} + u^{2} + u + 1)

u^{4} (u - 1) (u^{18} + u^{17} + u^{16} + u^{15} + u^{14} + u^{13} + u^{12} + u^{11} + u^{10} + u^{9} + u^{8} + u^{7} + u^{6} + u^{5} + u^{4} + u^{3} + u^{2} + u + 1) = 0

Sử dụng Nguyên tắc Hệ số 0: Nếu

ab = 0

thì

a = 0

b = 0

u = 0 or u - 1 = 0 or u^{18} + u^{17} + u^{16} + u^{15} + u^{14} + u^{13} + u^{12} + u^{11} + u^{10} + u^{9} + u^{8} + u^{7} + u^{6} + u^{5} + u^{4} + u^{3} + u^{2} + u + 1 = 0

Giải

u - 1 = 0 : u = 1

u - 1 = 0

Di chuyển

1

sang vế phải

u - 1 = 0

Thêm

1

vào cả hai bên

u - 1 + 1 = 0 + 1

Rút gọn

u = 1

u = 1

Giải

u^{18} + u^{17} + u^{16} + u^{15} + u^{14} + u^{13} + u^{12} + u^{11} + u^{10} + u^{9} + u^{8} + u^{7} + u^{6} + u^{5} + u^{4} + u^{3} + u^{2} + u + 1 = 0 :

Không có nghiệm cho

u \in R

u^{18} + u^{17} + u^{16} + u^{15} + u^{14} + u^{13} + u^{12} + u^{11} + u^{10} + u^{9} + u^{8} + u^{7} + u^{6} + u^{5} + u^{4} + u^{3} + u^{2} + u + 1 = 0

Tìm một lời giải cho

u^{18} + u^{17} + u^{16} + u^{15} + u^{14} + u^{13} + u^{12} + u^{11} + u^{10} + u^{9} + u^{8} + u^{7} + u^{6} + u^{5} + u^{4} + u^{3} + u^{2} + u + 1 = 0

bằng Newton-Raphson:

Không có nghiệm cho

u \in R

u^{18} + u^{17} + u^{16} + u^{15} + u^{14} + u^{13} + u^{12} + u^{11} + u^{10} + u^{9} + u^{8} + u^{7} + u^{6} + u^{5} + u^{4} + u^{3} + u^{2} + u + 1 = 0

Định nghĩa xấp xỉ Newton-Raphson

f (u) = u^{18} + u^{17} + u^{16} + u^{15} + u^{14} + u^{13} + u^{12} + u^{11} + u^{10} + u^{9} + u^{8} + u^{7} + u^{6} + u^{5} + u^{4} + u^{3} + u^{2} + u + 1

Tìm

f^{^{'}} (u) : 18 u^{17} + 17 u^{16} + 16 u^{15} + 15 u^{14} + 14 u^{13} + 13 u^{12} + 12 u^{11} + 11 u^{10} + 10 u^{9} + 9 u^{8} + 8 u^{7} + 7 u^{6} + 6 u^{5} + 5 u^{4} + 4 u^{3} + 3 u^{2} + 2 u + 1

\frac{d}{d u} (u^{18} + u^{17} + u^{16} + u^{15} + u^{14} + u^{13} + u^{12} + u^{11} + u^{10} + u^{9} + u^{8} + u^{7} + u^{6} + u^{5} + u^{4} + u^{3} + u^{2} + u + 1)

Áp dụng quy tắc Đạo hàm của một Tổng:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{18}) + \frac{d}{d u} (u^{17}) + \frac{d}{d u} (u^{16}) + \frac{d}{d u} (u^{15}) + \frac{d}{d u} (u^{14}) + \frac{d}{d u} (u^{13}) + \frac{d}{d u} (u^{12}) + \frac{d}{d u} (u^{11}) + \frac{d}{d u} (u^{10}) + \frac{d}{d u} (u^{9}) + \frac{d}{d u} (u^{8}) + \frac{d}{d u} (u^{7}) + \frac{d}{d u} (u^{6}) + \frac{d}{d u} (u^{5}) + \frac{d}{d u} (u^{4}) + \frac{d}{d u} (u^{3}) + \frac{d}{d u} (u^{2}) + \frac{d u}{d u} + \frac{d}{d u} (1)

\frac{d}{d u} (u^{18}) = 18 u^{17}

\frac{d}{d u} (u^{18})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 18 u^{18 - 1}

Rút gọn

= 18 u^{17}

\frac{d}{d u} (u^{17}) = 17 u^{16}

\frac{d}{d u} (u^{17})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 17 u^{17 - 1}

Rút gọn

= 17 u^{16}

\frac{d}{d u} (u^{16}) = 16 u^{15}

\frac{d}{d u} (u^{16})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 16 u^{16 - 1}

Rút gọn

= 16 u^{15}

\frac{d}{d u} (u^{15}) = 15 u^{14}

\frac{d}{d u} (u^{15})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 15 u^{15 - 1}

Rút gọn

= 15 u^{14}

\frac{d}{d u} (u^{14}) = 14 u^{13}

\frac{d}{d u} (u^{14})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 14 u^{14 - 1}

Rút gọn

= 14 u^{13}

\frac{d}{d u} (u^{13}) = 13 u^{12}

\frac{d}{d u} (u^{13})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 13 u^{13 - 1}

Rút gọn

= 13 u^{12}

\frac{d}{d u} (u^{12}) = 12 u^{11}

\frac{d}{d u} (u^{12})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 12 u^{12 - 1}

Rút gọn

= 12 u^{11}

\frac{d}{d u} (u^{11}) = 11 u^{10}

\frac{d}{d u} (u^{11})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 11 u^{11 - 1}

Rút gọn

= 11 u^{10}

\frac{d}{d u} (u^{10}) = 10 u^{9}

\frac{d}{d u} (u^{10})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 10 u^{10 - 1}

Rút gọn

= 10 u^{9}

\frac{d}{d u} (u^{9}) = 9 u^{8}

\frac{d}{d u} (u^{9})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 9 u^{9 - 1}

Rút gọn

= 9 u^{8}

\frac{d}{d u} (u^{8}) = 8 u^{7}

\frac{d}{d u} (u^{8})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 8 u^{8 - 1}

Rút gọn

= 8 u^{7}

\frac{d}{d u} (u^{7}) = 7 u^{6}

\frac{d}{d u} (u^{7})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 7 u^{7 - 1}

Rút gọn

= 7 u^{6}

\frac{d}{d u} (u^{6}) = 6 u^{5}

\frac{d}{d u} (u^{6})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 6 u^{6 - 1}

Rút gọn

= 6 u^{5}

\frac{d}{d u} (u^{5}) = 5 u^{4}

\frac{d}{d u} (u^{5})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 5 u^{5 - 1}

Rút gọn

= 5 u^{4}

\frac{d}{d u} (u^{4}) = 4 u^{3}

\frac{d}{d u} (u^{4})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 4 u^{4 - 1}

Rút gọn

= 4 u^{3}

\frac{d}{d u} (u^{3}) = 3 u^{2}

\frac{d}{d u} (u^{3})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 u^{3 - 1}

Rút gọn

= 3 u^{2}

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

Áp dụng Quy tắc Lũy thừa:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

Rút gọn

= 2 u

\frac{d u}{d u} = 1

\frac{d u}{d u}

Áp dụng đạo hàm chung:

\frac{d u}{d u} = 1

= 1

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

Đạo hàm của một hằng số:

\frac{d}{d x} (a) = 0

= 0

= 18 u^{17} + 17 u^{16} + 16 u^{15} + 15 u^{14} + 14 u^{13} + 13 u^{12} + 12 u^{11} + 11 u^{10} + 10 u^{9} + 9 u^{8} + 8 u^{7} + 7 u^{6} + 6 u^{5} + 5 u^{4} + 4 u^{3} + 3 u^{2} + 2 u + 1 + 0

Rút gọn

= 18 u^{17} + 17 u^{16} + 16 u^{15} + 15 u^{14} + 14 u^{13} + 13 u^{12} + 12 u^{11} + 11 u^{10} + 10 u^{9} + 9 u^{8} + 8 u^{7} + 7 u^{6} + 6 u^{5} + 5 u^{4} + 4 u^{3} + 3 u^{2} + 2 u + 1

Cho

u_{0} = - 1

Tính

u_{n + 1}

cho đến

Δ u_{n + 1} < 0.000001

u_{1} = - 0.88888 \dots : Δ u_{1} = 0.11111 \dots

f (u_{0}) = (- 1)^{18} + (- 1)^{17} + (- 1)^{16} + (- 1)^{15} + (- 1)^{14} + (- 1)^{13} + (- 1)^{12} + (- 1)^{11} + (- 1)^{10} + (- 1)^{9} + (- 1)^{8} + (- 1)^{7} + (- 1)^{6} + (- 1)^{5} + (- 1)^{4} + (- 1)^{3} + (- 1)^{2} + (- 1) + 1 = 1

f^{^{'}} (u_{0}) = 18 (- 1)^{17} + 17 (- 1)^{16} + 16 (- 1)^{15} + 15 (- 1)^{14} + 14 (- 1)^{13} + 13 (- 1)^{12} + 12 (- 1)^{11} + 11 (- 1)^{10} + 10 (- 1)^{9} + 9 (- 1)^{8} + 8 (- 1)^{7} + 7 (- 1)^{6} + 6 (- 1)^{5} + 5 (- 1)^{4} + 4 (- 1)^{3} + 3 (- 1)^{2} + 2 (- 1) + 1 = - 9

u_{1} = - 0.88888 \dots

Δ u_{1} = ∣ - 0.88888 \dots - (- 1) ∣ = 0.11111 \dots

Δ u_{1} = 0.11111 \dots

u_{2} = - 0.23578 \dots : Δ u_{2} = 0.65310 \dots

f (u_{1}) = (- 0.88888 \dots)^{18} + (- 0.88888 \dots)^{17} + (- 0.88888 \dots)^{16} + (- 0.88888 \dots)^{15} + (- 0.88888 \dots)^{14} + (- 0.88888 \dots)^{13} + (- 0.88888 \dots)^{12} + (- 0.88888 \dots)^{11} + (- 0.88888 \dots)^{10} + (- 0.88888 \dots)^{9} + (- 0.88888 \dots)^{8} + (- 0.88888 \dots)^{7} + (- 0.88888 \dots)^{6} + (- 0.88888 \dots)^{5} + (- 0.88888 \dots)^{4} + (- 0.88888 \dots)^{3} + (- 0.88888 \dots)^{2} + (- 0.88888 \dots) + 1 = 0.58589 \dots

f^{^{'}} (u_{1}) = 18 (- 0.88888 \dots)^{17} + 17 (- 0.88888 \dots)^{16} + 16 (- 0.88888 \dots)^{15} + 15 (- 0.88888 \dots)^{14} + 14 (- 0.88888 \dots)^{13} + 13 (- 0.88888 \dots)^{12} + 12 (- 0.88888 \dots)^{11} + 11 (- 0.88888 \dots)^{10} + 10 (- 0.88888 \dots)^{9} + 9 (- 0.88888 \dots)^{8} + 8 (- 0.88888 \dots)^{7} + 7 (- 0.88888 \dots)^{6} + 6 (- 0.88888 \dots)^{5} + 5 (- 0.88888 \dots)^{4} + 4 (- 0.88888 \dots)^{3} + 3 (- 0.88888 \dots)^{2} + 2 (- 0.88888 \dots) + 1 = - 0.89708 \dots

u_{2} = - 0.23578 \dots

Δ u_{2} = ∣ - 0.23578 \dots - (- 0.88888 \dots) ∣ = 0.65310 \dots

Δ u_{2} = 0.65310 \dots

u_{3} = - 1.47156 \dots : Δ u_{3} = 1.23578 \dots

f (u_{2}) = (- 0.23578 \dots)^{18} + (- 0.23578 \dots)^{17} + (- 0.23578 \dots)^{16} + (- 0.23578 \dots)^{15} + (- 0.23578 \dots)^{14} + (- 0.23578 \dots)^{13} + (- 0.23578 \dots)^{12} + (- 0.23578 \dots)^{11} + (- 0.23578 \dots)^{10} + (- 0.23578 \dots)^{9} + (- 0.23578 \dots)^{8} + (- 0.23578 \dots)^{7} + (- 0.23578 \dots)^{6} + (- 0.23578 \dots)^{5} + (- 0.23578 \dots)^{4} + (- 0.23578 \dots)^{3} + (- 0.23578 \dots)^{2} + (- 0.23578 \dots) + 1 = 0.80920 \dots

f^{^{'}} (u_{2}) = 18 (- 0.23578 \dots)^{17} + 17 (- 0.23578 \dots)^{16} + 16 (- 0.23578 \dots)^{15} + 15 (- 0.23578 \dots)^{14} + 14 (- 0.23578 \dots)^{13} + 13 (- 0.23578 \dots)^{12} + 12 (- 0.23578 \dots)^{11} + 11 (- 0.23578 \dots)^{10} + 10 (- 0.23578 \dots)^{9} + 9 (- 0.23578 \dots)^{8} + 8 (- 0.23578 \dots)^{7} + 7 (- 0.23578 \dots)^{6} + 6 (- 0.23578 \dots)^{5} + 5 (- 0.23578 \dots)^{4} + 4 (- 0.23578 \dots)^{3} + 3 (- 0.23578 \dots)^{2} + 2 (- 0.23578 \dots) + 1 = 0.65481 \dots

u_{3} = - 1.47156 \dots

Δ u_{3} = ∣ - 1.47156 \dots - (- 0.23578 \dots) ∣ = 1.23578 \dots

Δ u_{3} = 1.23578 \dots

u_{4} = - 1.39155 \dots : Δ u_{4} = 0.08000 \dots

f (u_{3}) = (- 1.47156 \dots)^{18} + (- 1.47156 \dots)^{17} + (- 1.47156 \dots)^{16} + (- 1.47156 \dots)^{15} + (- 1.47156 \dots)^{14} + (- 1.47156 \dots)^{13} + (- 1.47156 \dots)^{12} + (- 1.47156 \dots)^{11} + (- 1.47156 \dots)^{10} + (- 1.47156 \dots)^{9} + (- 1.47156 \dots)^{8} + (- 1.47156 \dots)^{7} + (- 1.47156 \dots)^{6} + (- 1.47156 \dots)^{5} + (- 1.47156 \dots)^{4} + (- 1.47156 \dots)^{3} + (- 1.47156 \dots)^{2} + (- 1.47156 \dots) + 1 = 623.90302 \dots

f^{^{'}} (u_{3}) = 18 (- 1.47156 \dots)^{17} + 17 (- 1.47156 \dots)^{16} + 16 (- 1.47156 \dots)^{15} + 15 (- 1.47156 \dots)^{14} + 14 (- 1.47156 \dots)^{13} + 13 (- 1.47156 \dots)^{12} + 12 (- 1.47156 \dots)^{11} + 11 (- 1.47156 \dots)^{10} + 10 (- 1.47156 \dots)^{9} + 9 (- 1.47156 \dots)^{8} + 8 (- 1.47156 \dots)^{7} + 7 (- 1.47156 \dots)^{6} + 6 (- 1.47156 \dots)^{5} + 5 (- 1.47156 \dots)^{4} + 4 (- 1.47156 \dots)^{3} + 3 (- 1.47156 \dots)^{2} + 2 (- 1.47156 \dots) + 1 = - 7797.82245 \dots

u_{4} = - 1.39155 \dots

Δ u_{4} = ∣ - 1.39155 \dots - (- 1.47156 \dots) ∣ = 0.08000 \dots

Δ u_{4} = 0.08000 \dots

u_{5} = - 1.31585 \dots : Δ u_{5} = 0.07569 \dots

f (u_{4}) = (- 1.39155 \dots)^{18} + (- 1.39155 \dots)^{17} + (- 1.39155 \dots)^{16} + (- 1.39155 \dots)^{15} + (- 1.39155 \dots)^{14} + (- 1.39155 \dots)^{13} + (- 1.39155 \dots)^{12} + (- 1.39155 \dots)^{11} + (- 1.39155 \dots)^{10} + (- 1.39155 \dots)^{9} + (- 1.39155 \dots)^{8} + (- 1.39155 \dots)^{7} + (- 1.39155 \dots)^{6} + (- 1.39155 \dots)^{5} + (- 1.39155 \dots)^{4} + (- 1.39155 \dots)^{3} + (- 1.39155 \dots)^{2} + (- 1.39155 \dots) + 1 = 223.17190 \dots

f^{^{'}} (u_{4}) = 18 (- 1.39155 \dots)^{17} + 17 (- 1.39155 \dots)^{16} + 16 (- 1.39155 \dots)^{15} + 15 (- 1.39155 \dots)^{14} + 14 (- 1.39155 \dots)^{13} + 13 (- 1.39155 \dots)^{12} + 12 (- 1.39155 \dots)^{11} + 11 (- 1.39155 \dots)^{10} + 10 (- 1.39155 \dots)^{9} + 9 (- 1.39155 \dots)^{8} + 8 (- 1.39155 \dots)^{7} + 7 (- 1.39155 \dots)^{6} + 6 (- 1.39155 \dots)^{5} + 5 (- 1.39155 \dots)^{4} + 4 (- 1.39155 \dots)^{3} + 3 (- 1.39155 \dots)^{2} + 2 (- 1.39155 \dots) + 1 = - 2948.11712 \dots

u_{5} = - 1.31585 \dots

Δ u_{5} = ∣ - 1.31585 \dots - (- 1.39155 \dots) ∣ = 0.07569 \dots

Δ u_{5} = 0.07569 \dots

u_{6} = - 1.24406 \dots : Δ u_{6} = 0.07179 \dots

f (u_{5}) = (- 1.31585 \dots)^{18} + (- 1.31585 \dots)^{17} + (- 1.31585 \dots)^{16} + (- 1.31585 \dots)^{15} + (- 1.31585 \dots)^{14} + (- 1.31585 \dots)^{13} + (- 1.31585 \dots)^{12} + (- 1.31585 \dots)^{11} + (- 1.31585 \dots)^{10} + (- 1.31585 \dots)^{9} + (- 1.31585 \dots)^{8} + (- 1.31585 \dots)^{7} + (- 1.31585 \dots)^{6} + (- 1.31585 \dots)^{5} + (- 1.31585 \dots)^{4} + (- 1.31585 \dots)^{3} + (- 1.31585 \dots)^{2} + (- 1.31585 \dots) + 1 = 79.90865 \dots

f^{^{'}} (u_{5}) = 18 (- 1.31585 \dots)^{17} + 17 (- 1.31585 \dots)^{16} + 16 (- 1.31585 \dots)^{15} + 15 (- 1.31585 \dots)^{14} + 14 (- 1.31585 \dots)^{13} + 13 (- 1.31585 \dots)^{12} + 12 (- 1.31585 \dots)^{11} + 11 (- 1.31585 \dots)^{10} + 10 (- 1.31585 \dots)^{9} + 9 (- 1.31585 \dots)^{8} + 8 (- 1.31585 \dots)^{7} + 7 (- 1.31585 \dots)^{6} + 6 (- 1.31585 \dots)^{5} + 5 (- 1.31585 \dots)^{4} + 4 (- 1.31585 \dots)^{3} + 3 (- 1.31585 \dots)^{2} + 2 (- 1.31585 \dots) + 1 = - 1113.08361 \dots

u_{6} = - 1.24406 \dots

Δ u_{6} = ∣ - 1.24406 \dots - (- 1.31585 \dots) ∣ = 0.07179 \dots

Δ u_{6} = 0.07179 \dots

u_{7} = - 1.17552 \dots : Δ u_{7} = 0.06854 \dots

f (u_{6}) = (- 1.24406 \dots)^{18} + (- 1.24406 \dots)^{17} + (- 1.24406 \dots)^{16} + (- 1.24406 \dots)^{15} + (- 1.24406 \dots)^{14} + (- 1.24406 \dots)^{13} + (- 1.24406 \dots)^{12} + (- 1.24406 \dots)^{11} + (- 1.24406 \dots)^{10} + (- 1.24406 \dots)^{9} + (- 1.24406 \dots)^{8} + (- 1.24406 \dots)^{7} + (- 1.24406 \dots)^{6} + (- 1.24406 \dots)^{5} + (- 1.24406 \dots)^{4} + (- 1.24406 \dots)^{3} + (- 1.24406 \dots)^{2} + (- 1.24406 \dots) + 1 = 28.69312 \dots

f^{^{'}} (u_{6}) = 18 (- 1.24406 \dots)^{17} + 17 (- 1.24406 \dots)^{16} + 16 (- 1.24406 \dots)^{15} + 15 (- 1.24406 \dots)^{14} + 14 (- 1.24406 \dots)^{13} + 13 (- 1.24406 \dots)^{12} + 12 (- 1.24406 \dots)^{11} + 11 (- 1.24406 \dots)^{10} + 10 (- 1.24406 \dots)^{9} + 9 (- 1.24406 \dots)^{8} + 8 (- 1.24406 \dots)^{7} + 7 (- 1.24406 \dots)^{6} + 6 (- 1.24406 \dots)^{5} + 5 (- 1.24406 \dots)^{4} + 4 (- 1.24406 \dots)^{3} + 3 (- 1.24406 \dots)^{2} + 2 (- 1.24406 \dots) + 1 = - 418.62427 \dots

u_{7} = - 1.17552 \dots

Δ u_{7} = ∣ - 1.17552 \dots - (- 1.24406 \dots) ∣ = 0.06854 \dots

Δ u_{7} = 0.06854 \dots

u_{8} = - 1.10880 \dots : Δ u_{8} = 0.06671 \dots

f (u_{7}) = (- 1.17552 \dots)^{18} + (- 1.17552 \dots)^{17} + (- 1.17552 \dots)^{16} + (- 1.17552 \dots)^{15} + (- 1.17552 \dots)^{14} + (- 1.17552 \dots)^{13} + (- 1.17552 \dots)^{12} + (- 1.17552 \dots)^{11} + (- 1.17552 \dots)^{10} + (- 1.17552 \dots)^{9} + (- 1.17552 \dots)^{8} + (- 1.17552 \dots)^{7} + (- 1.17552 \dots)^{6} + (- 1.17552 \dots)^{5} + (- 1.17552 \dots)^{4} + (- 1.17552 \dots)^{3} + (- 1.17552 \dots)^{2} + (- 1.17552 \dots) + 1 = 10.38689 \dots

f^{^{'}} (u_{7}) = 18 (- 1.17552 \dots)^{17} + 17 (- 1.17552 \dots)^{16} + 16 (- 1.17552 \dots)^{15} + 15 (- 1.17552 \dots)^{14} + 14 (- 1.17552 \dots)^{13} + 13 (- 1.17552 \dots)^{12} + 12 (- 1.17552 \dots)^{11} + 11 (- 1.17552 \dots)^{10} + 10 (- 1.17552 \dots)^{9} + 9 (- 1.17552 \dots)^{8} + 8 (- 1.17552 \dots)^{7} + 7 (- 1.17552 \dots)^{6} + 6 (- 1.17552 \dots)^{5} + 5 (- 1.17552 \dots)^{4} + 4 (- 1.17552 \dots)^{3} + 3 (- 1.17552 \dots)^{2} + 2 (- 1.17552 \dots) + 1 = - 155.67966 \dots

u_{8} = - 1.10880 \dots

Δ u_{8} = ∣ - 1.10880 \dots - (- 1.17552 \dots) ∣ = 0.06671 \dots

Δ u_{8} = 0.06671 \dots

u_{9} = - 1.04007 \dots : Δ u_{9} = 0.06872 \dots

f (u_{8}) = (- 1.10880 \dots)^{18} + (- 1.10880 \dots)^{17} + (- 1.10880 \dots)^{16} + (- 1.10880 \dots)^{15} + (- 1.10880 \dots)^{14} + (- 1.10880 \dots)^{13} + (- 1.10880 \dots)^{12} + (- 1.10880 \dots)^{11} + (- 1.10880 \dots)^{10} + (- 1.10880 \dots)^{9} + (- 1.10880 \dots)^{8} + (- 1.10880 \dots)^{7} + (- 1.10880 \dots)^{6} + (- 1.10880 \dots)^{5} + (- 1.10880 \dots)^{4} + (- 1.10880 \dots)^{3} + (- 1.10880 \dots)^{2} + (- 1.10880 \dots) + 1 = 3.84863 \dots

f^{^{'}} (u_{8}) = 18 (- 1.10880 \dots)^{17} + 17 (- 1.10880 \dots)^{16} + 16 (- 1.10880 \dots)^{15} + 15 (- 1.10880 \dots)^{14} + 14 (- 1.10880 \dots)^{13} + 13 (- 1.10880 \dots)^{12} + 12 (- 1.10880 \dots)^{11} + 11 (- 1.10880 \dots)^{10} + 10 (- 1.10880 \dots)^{9} + 9 (- 1.10880 \dots)^{8} + 8 (- 1.10880 \dots)^{7} + 7 (- 1.10880 \dots)^{6} + 6 (- 1.10880 \dots)^{5} + 5 (- 1.10880 \dots)^{4} + 4 (- 1.10880 \dots)^{3} + 3 (- 1.10880 \dots)^{2} + 2 (- 1.10880 \dots) + 1 = - 55.99781 \dots

u_{9} = - 1.04007 \dots

Δ u_{9} = ∣ - 1.04007 \dots - (- 1.10880 \dots) ∣ = 0.06872 \dots

Δ u_{9} = 0.06872 \dots

u_{10} = - 0.95606 \dots : Δ u_{10} = 0.08401 \dots

f (u_{9}) = (- 1.04007 \dots)^{18} + (- 1.04007 \dots)^{17} + (- 1.04007 \dots)^{16} + (- 1.04007 \dots)^{15} + (- 1.04007 \dots)^{14} + (- 1.04007 \dots)^{13} + (- 1.04007 \dots)^{12} + (- 1.04007 \dots)^{11} + (- 1.04007 \dots)^{10} + (- 1.04007 \dots)^{9} + (- 1.04007 \dots)^{8} + (- 1.04007 \dots)^{7} + (- 1.04007 \dots)^{6} + (- 1.04007 \dots)^{5} + (- 1.04007 \dots)^{4} + (- 1.04007 \dots)^{3} + (- 1.04007 \dots)^{2} + (- 1.04007 \dots) + 1 = 1.52432 \dots

f^{^{'}} (u_{9}) = 18 (- 1.04007 \dots)^{17} + 17 (- 1.04007 \dots)^{16} + 16 (- 1.04007 \dots)^{15} + 15 (- 1.04007 \dots)^{14} + 14 (- 1.04007 \dots)^{13} + 13 (- 1.04007 \dots)^{12} + 12 (- 1.04007 \dots)^{11} + 11 (- 1.04007 \dots)^{10} + 10 (- 1.04007 \dots)^{9} + 9 (- 1.04007 \dots)^{8} + 8 (- 1.04007 \dots)^{7} + 7 (- 1.04007 \dots)^{6} + 6 (- 1.04007 \dots)^{5} + 5 (- 1.04007 \dots)^{4} + 4 (- 1.04007 \dots)^{3} + 3 (- 1.04007 \dots)^{2} + 2 (- 1.04007 \dots) + 1 = - 18.14456 \dots

u_{10} = - 0.95606 \dots

Δ u_{10} = ∣ - 0.95606 \dots - (- 1.04007 \dots) ∣ = 0.08401 \dots

Δ u_{10} = 0.08401 \dots

Không thể tìm được lời giải

Giải pháp là

Kh \overset{o}{^} ng c \overset{o}{ˊ} nghi ệ m cho u \in R

Các lời giải là

u = 0, u = 1

Thay thế lại

u = cos (x)

cos (x) = 0, cos (x) = 1

cos (x) = 0, cos (x) = 1

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

Các lời giải chung cho

cos (x) = 0

cos (x)

bảng tuần hoàn với chu kỳ

2 πn

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 1 : x = 2 πn

cos (x) = 1

Các lời giải chung cho

cos (x) = 1

cos (x)

bảng tuần hoàn với chu kỳ

2 πn

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 2 πn

Giải

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Kết hợp tất cả các cách giải

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2 πn

Đồ Thị

Ví dụ phổ biến

cos^4(x)+2cos^2(x)=1

cos^{4} (x) + 2 cos^{2} (x) = 1

cos^2(x)+sin^2(x)=cos^5(x)

cos^{2} (x) + sin^{2} (x) = cos^{5} (x)

sin(x-45^5)=((sqrt(2)))/2

sin (x - 4 5^{5}) = \frac{( 2 )}{2}

(sin(x)-sqrt(3)*cos(x))/2 =0

\frac{sin ( x ) - 3 \cdot cos ( x )}{2} = 0

cos(1/(3x))= 1/3

cos (\frac{1}{3 x}) = \frac{1}{3}