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受欢迎的三角函数 >

sin^4(x)=-cos(x)

解答

sin^{4} (x) = - cos (x)

解答

x = π - 1.01821 \dots + 2 πn, x = π + 1.01821 \dots + 2 πn

+1

度数

x = 121.66074 \dots^{\circ} + 36 0^{\circ} n, x = 238.33925 \dots^{\circ} + 36 0^{\circ} n

求解步骤

sin^{4} (x) = - cos (x)

两边进行平方

(sin^{4} (x))^{2} = (- cos (x))^{2}

两边减去

(- cos (x))^{2}

sin^{8} (x) - cos^{2} (x) = 0

使用三角恒等式改写

- cos^{2} (x) + sin^{8} (x)

使用毕达哥拉斯恒等式:

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

cos^{2} (x) = 1 - sin^{2} (x)

= - (1 - sin^{2} (x)) + sin^{8} (x)

- (1 - sin^{2} (x)) : - 1 + sin^{2} (x)

- (1 - sin^{2} (x))

打开括号

= - (1) - (- sin^{2} (x))

使用加减运算法则

- (- a) = a, - (a) = - a

= - 1 + sin^{2} (x)

= - 1 + sin^{2} (x) + sin^{8} (x)

- 1 + sin^{2} (x) + sin^{8} (x) = 0

用替代法求解

- 1 + sin^{2} (x) + sin^{8} (x) = 0

令

: sin (x) = u

- 1 + u^{2} + u^{8} = 0

- 1 + u^{2} + u^{8} = 0 : u = 0.72449 \dots, u = - 0.72449 \dots

- 1 + u^{2} + u^{8} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{8} + u^{2} - 1 = 0

用

v = u^{2}

和

v^{4} = u^{8}

改写方程式

v^{4} + v - 1 = 0

解

v^{4} + v - 1 = 0 : v \approx 0.72449 \dots, v \approx - 1.22074 \dots

v^{4} + v - 1 = 0

使用牛顿-拉弗森方法找到

v^{4} + v - 1 = 0

的一个解:

v \approx 0.72449 \dots

v^{4} + v - 1 = 0

牛顿-拉弗森近似法定义

f (v) = v^{4} + v - 1

找到

f^{^{'}} (v) : 4 v^{3} + 1

\frac{d}{d v} (v^{4} + v - 1)

使用微分加减法定则:

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

= \frac{d}{d v} (v^{4}) + \frac{d v}{d v} - \frac{d}{d v} (1)

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

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

使用幂法则:

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

= 4 v^{4 - 1}

化简

= 4 v^{3}

\frac{d v}{d v} = 1

\frac{d v}{d v}

使用常见微分定则:

\frac{d v}{d v} = 1

= 1

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

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

常数微分:

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

= 0

= 4 v^{3} + 1 - 0

化简

= 4 v^{3} + 1

令

v_{0} = 1

计算

v_{n + 1}

至

Δ v_{n + 1} < 0.000001

v_{1} = 0.8 : Δ v_{1} = 0.2

f (v_{0}) = 1^{4} + 1 - 1 = 1

f^{^{'}} (v_{0}) = 4 \cdot 1^{3} + 1 = 5

v_{1} = 0.8

Δ v_{1} = ∣ 0.8 - 1 ∣ = 0.2

Δ v_{1} = 0.2

v_{2} = 0.73123 \dots : Δ v_{2} = 0.06876 \dots

f (v_{1}) = 0. 8^{4} + 0.8 - 1 = 0.2096

f^{^{'}} (v_{1}) = 4 \cdot 0. 8^{3} + 1 = 3.048

v_{2} = 0.73123 \dots

Δ v_{2} = ∣ 0.73123 \dots - 0.8 ∣ = 0.06876 \dots

Δ v_{2} = 0.06876 \dots

v_{3} = 0.72454 \dots : Δ v_{3} = 0.00668 \dots

f (v_{2}) = 0.73123 \dots^{4} + 0.73123 \dots - 1 = 0.01714 \dots

f^{^{'}} (v_{2}) = 4 \cdot 0.73123 \dots^{3} + 1 = 2.56396 \dots

v_{3} = 0.72454 \dots

Δ v_{3} = ∣ 0.72454 \dots - 0.73123 \dots ∣ = 0.00668 \dots

Δ v_{3} = 0.00668 \dots

v_{4} = 0.72449 \dots : Δ v_{4} = 0.00005 \dots

f (v_{3}) = 0.72454 \dots^{4} + 0.72454 \dots - 1 = 0.00014 \dots

f^{^{'}} (v_{3}) = 4 \cdot 0.72454 \dots^{3} + 1 = 2.52146 \dots

v_{4} = 0.72449 \dots

Δ v_{4} = ∣ 0.72449 \dots - 0.72454 \dots ∣ = 0.00005 \dots

Δ v_{4} = 0.00005 \dots

v_{5} = 0.72449 \dots : Δ v_{5} = 3.99053 E - 9

f (v_{4}) = 0.72449 \dots^{4} + 0.72449 \dots - 1 = 1.00606 E - 8

f^{^{'}} (v_{4}) = 4 \cdot 0.72449 \dots^{3} + 1 = 2.52111 \dots

v_{5} = 0.72449 \dots

Δ v_{5} = ∣ 0.72449 \dots - 0.72449 \dots ∣ = 3.99053 E - 9

Δ v_{5} = 3.99053 E - 9

v \approx 0.72449 \dots

使用长除法

Eq u a t i o n 0 : \frac{v ^{4} + v - 1}{v - 0.72449 \dots} = v^{3} + 0.72449 \dots v^{2} + 0.52488 \dots v + 1.38027 \dots

v^{3} + 0.72449 \dots v^{2} + 0.52488 \dots v + 1.38027 \dots \approx 0

使用牛顿-拉弗森方法找到

v^{3} + 0.72449 \dots v^{2} + 0.52488 \dots v + 1.38027 \dots = 0

的一个解:

v \approx - 1.22074 \dots

v^{3} + 0.72449 \dots v^{2} + 0.52488 \dots v + 1.38027 \dots = 0

牛顿-拉弗森近似法定义

f (v) = v^{3} + 0.72449 \dots v^{2} + 0.52488 \dots v + 1.38027 \dots

找到

f^{^{'}} (v) : 3 v^{2} + 1.44898 \dots v + 0.52488 \dots

\frac{d}{d v} (v^{3} + 0.72449 \dots v^{2} + 0.52488 \dots v + 1.38027 \dots)

使用微分加减法定则:

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

= \frac{d}{d v} (v^{3}) + \frac{d}{d v} (0.72449 \dots v^{2}) + \frac{d}{d v} (0.52488 \dots v) + \frac{d}{d v} (1.38027 \dots)

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

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

使用幂法则:

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

= 3 v^{3 - 1}

化简

= 3 v^{2}

\frac{d}{d v} (0.72449 \dots v^{2}) = 1.44898 \dots v

\frac{d}{d v} (0.72449 \dots v^{2})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 0.72449 \dots \frac{d}{d v} (v^{2})

使用幂法则:

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

= 0.72449 \dots \cdot 2 v^{2 - 1}

化简

= 1.44898 \dots v

\frac{d}{d v} (0.52488 \dots v) = 0.52488 \dots

\frac{d}{d v} (0.52488 \dots v)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 0.52488 \dots \frac{d v}{d v}

使用常见微分定则:

\frac{d v}{d v} = 1

= 0.52488 \dots \cdot 1

化简

= 0.52488 \dots

\frac{d}{d v} (1.38027 \dots) = 0

\frac{d}{d v} (1.38027 \dots)

常数微分:

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

= 0

= 3 v^{2} + 1.44898 \dots v + 0.52488 \dots + 0

化简

= 3 v^{2} + 1.44898 \dots v + 0.52488 \dots

令

v_{0} = - 3

计算

v_{n + 1}

至

Δ v_{n + 1} < 0.000001

v_{1} = - 2.10803 \dots : Δ v_{1} = 0.89196 \dots

f (v_{0}) = (- 3)^{3} + 0.72449 \dots (- 3)^{2} + 0.52488 \dots (- 3) + 1.38027 \dots = - 20.67396 \dots

f^{^{'}} (v_{0}) = 3 (- 3)^{2} + 1.44898 \dots (- 3) + 0.52488 \dots = 23.17793 \dots

v_{1} = - 2.10803 \dots

Δ v_{1} = ∣ - 2.10803 \dots - (- 3) ∣ = 0.89196 \dots

Δ v_{1} = 0.89196 \dots

v_{2} = - 1.56419 \dots : Δ v_{2} = 0.54383 \dots

f (v_{1}) = (- 2.10803 \dots)^{3} + 0.72449 \dots (- 2.10803 \dots)^{2} + 0.52488 \dots (- 2.10803 \dots) + 1.38027 \dots = - 5.87438 \dots

f^{^{'}} (v_{1}) = 3 (- 2.10803 \dots)^{2} + 1.44898 \dots (- 2.10803 \dots) + 0.52488 \dots = 10.80178 \dots

v_{2} = - 1.56419 \dots

Δ v_{2} = ∣ - 1.56419 \dots - (- 2.10803 \dots) ∣ = 0.54383 \dots

Δ v_{2} = 0.54383 \dots

v_{3} = - 1.29711 \dots : Δ v_{3} = 0.26708 \dots

f (v_{2}) = (- 1.56419 \dots)^{3} + 0.72449 \dots (- 1.56419 \dots)^{2} + 0.52488 \dots (- 1.56419 \dots) + 1.38027 \dots = - 1.49527 \dots

f^{^{'}} (v_{2}) = 3 (- 1.56419 \dots)^{2} + 1.44898 \dots (- 1.56419 \dots) + 0.52488 \dots = 5.59853 \dots

v_{3} = - 1.29711 \dots

Δ v_{3} = ∣ - 1.29711 \dots - (- 1.56419 \dots) ∣ = 0.26708 \dots

Δ v_{3} = 0.26708 \dots

v_{4} = - 1.22562 \dots : Δ v_{4} = 0.07148 \dots

f (v_{3}) = (- 1.29711 \dots)^{3} + 0.72449 \dots (- 1.29711 \dots)^{2} + 0.52488 \dots (- 1.29711 \dots) + 1.38027 \dots = - 0.26400 \dots

f^{^{'}} (v_{3}) = 3 (- 1.29711 \dots)^{2} + 1.44898 \dots (- 1.29711 \dots) + 0.52488 \dots = 3.69291 \dots

v_{4} = - 1.22562 \dots

Δ v_{4} = ∣ - 1.22562 \dots - (- 1.29711 \dots) ∣ = 0.07148 \dots

Δ v_{4} = 0.07148 \dots

v_{5} = - 1.22076 \dots : Δ v_{5} = 0.00485 \dots

f (v_{4}) = (- 1.22562 \dots)^{3} + 0.72449 \dots (- 1.22562 \dots)^{2} + 0.52488 \dots (- 1.22562 \dots) + 1.38027 \dots = - 0.01581 \dots

f^{^{'}} (v_{4}) = 3 (- 1.22562 \dots)^{2} + 1.44898 \dots (- 1.22562 \dots) + 0.52488 \dots = 3.25544 \dots

v_{5} = - 1.22076 \dots

Δ v_{5} = ∣ - 1.22076 \dots - (- 1.22562 \dots) ∣ = 0.00485 \dots

Δ v_{5} = 0.00485 \dots

v_{6} = - 1.22074 \dots : Δ v_{6} = 0.00002 \dots

f (v_{5}) = (- 1.22076 \dots)^{3} + 0.72449 \dots (- 1.22076 \dots)^{2} + 0.52488 \dots (- 1.22076 \dots) + 1.38027 \dots = - 0.00006 \dots

f^{^{'}} (v_{5}) = 3 (- 1.22076 \dots)^{2} + 1.44898 \dots (- 1.22076 \dots) + 0.52488 \dots = 3.22682 \dots

v_{6} = - 1.22074 \dots

Δ v_{6} = ∣ - 1.22074 \dots - (- 1.22076 \dots) ∣ = 0.00002 \dots

Δ v_{6} = 0.00002 \dots

v_{7} = - 1.22074 \dots : Δ v_{7} = 4.23633 E - 10

f (v_{6}) = (- 1.22074 \dots)^{3} + 0.72449 \dots (- 1.22074 \dots)^{2} + 0.52488 \dots (- 1.22074 \dots) + 1.38027 \dots = - 1.36693 E - 9

f^{^{'}} (v_{6}) = 3 (- 1.22074 \dots)^{2} + 1.44898 \dots (- 1.22074 \dots) + 0.52488 \dots = 3.22669 \dots

v_{7} = - 1.22074 \dots

Δ v_{7} = ∣ - 1.22074 \dots - (- 1.22074 \dots) ∣ = 4.23633 E - 10

Δ v_{7} = 4.23633 E - 10

v \approx - 1.22074 \dots

使用长除法

Eq u a t i o n 0 : \frac{v ^{3} + 0.72449 \dots v ^{2} + 0.52488 \dots v + 1.38027 \dots}{v + 1.22074 \dots} = v^{2} - 0.49625 \dots v + 1.13068 \dots

v^{2} - 0.49625 \dots v + 1.13068 \dots \approx 0

使用牛顿-拉弗森方法找到

v^{2} - 0.49625 \dots v + 1.13068 \dots = 0

的一个解:

v \in R

无解

v^{2} - 0.49625 \dots v + 1.13068 \dots = 0

牛顿-拉弗森近似法定义

f (v) = v^{2} - 0.49625 \dots v + 1.13068 \dots

找到

f^{^{'}} (v) : 2 v - 0.49625 \dots

\frac{d}{d v} (v^{2} - 0.49625 \dots v + 1.13068 \dots)

使用微分加减法定则:

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

= \frac{d}{d v} (v^{2}) - \frac{d}{d v} (0.49625 \dots v) + \frac{d}{d v} (1.13068 \dots)

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

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

使用幂法则:

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

= 2 v^{2 - 1}

化简

= 2 v

\frac{d}{d v} (0.49625 \dots v) = 0.49625 \dots

\frac{d}{d v} (0.49625 \dots v)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 0.49625 \dots \frac{d v}{d v}

使用常见微分定则:

\frac{d v}{d v} = 1

= 0.49625 \dots \cdot 1

化简

= 0.49625 \dots

\frac{d}{d v} (1.13068 \dots) = 0

\frac{d}{d v} (1.13068 \dots)

常数微分:

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

= 0

= 2 v - 0.49625 \dots + 0

化简

= 2 v - 0.49625 \dots

令

v_{0} = 2

计算

v_{n + 1}

至

Δ v_{n + 1} < 0.000001

v_{1} = 0.81892 \dots : Δ v_{1} = 1.18107 \dots

f (v_{0}) = 2^{2} - 0.49625 \dots \cdot 2 + 1.13068 \dots = 4.13818 \dots

f^{^{'}} (v_{0}) = 2 \cdot 2 - 0.49625 \dots = 3.50374 \dots

v_{1} = 0.81892 \dots

Δ v_{1} = ∣ 0.81892 \dots - 2 ∣ = 1.18107 \dots

Δ v_{1} = 1.18107 \dots

v_{2} = - 0.40298 \dots : Δ v_{2} = 1.22190 \dots

f (v_{1}) = 0.81892 \dots^{2} - 0.49625 \dots \cdot 0.81892 \dots + 1.13068 \dots = 1.39493 \dots

f^{^{'}} (v_{1}) = 2 \cdot 0.81892 \dots - 0.49625 \dots = 1.14160 \dots

v_{2} = - 0.40298 \dots

Δ v_{2} = ∣ - 0.40298 \dots - 0.81892 \dots ∣ = 1.22190 \dots

Δ v_{2} = 1.22190 \dots

v_{3} = 0.74357 \dots : Δ v_{3} = 1.14655 \dots

f (v_{2}) = (- 0.40298 \dots)^{2} - 0.49625 \dots (- 0.40298 \dots) + 1.13068 \dots = 1.49305 \dots

f^{^{'}} (v_{2}) = 2 (- 0.40298 \dots) - 0.49625 \dots = - 1.30221 \dots

v_{3} = 0.74357 \dots

Δ v_{3} = ∣ 0.74357 \dots - (- 0.40298 \dots) ∣ = 1.14655 \dots

Δ v_{3} = 1.14655 \dots

v_{4} = - 0.58309 \dots : Δ v_{4} = 1.32666 \dots

f (v_{3}) = 0.74357 \dots^{2} - 0.49625 \dots \cdot 0.74357 \dots + 1.13068 \dots = 1.31458 \dots

f^{^{'}} (v_{3}) = 2 \cdot 0.74357 \dots - 0.49625 \dots = 0.99089 \dots

v_{4} = - 0.58309 \dots

Δ v_{4} = ∣ - 0.58309 \dots - 0.74357 \dots ∣ = 1.32666 \dots

Δ v_{4} = 1.32666 \dots

v_{5} = 0.47562 \dots : Δ v_{5} = 1.05871 \dots

f (v_{4}) = (- 0.58309 \dots)^{2} - 0.49625 \dots (- 0.58309 \dots) + 1.13068 \dots = 1.76004 \dots

f^{^{'}} (v_{4}) = 2 (- 0.58309 \dots) - 0.49625 \dots = - 1.66243 \dots

v_{5} = 0.47562 \dots

Δ v_{5} = ∣ 0.47562 \dots - (- 0.58309 \dots) ∣ = 1.05871 \dots

Δ v_{5} = 1.05871 \dots

v_{6} = - 1.98788 \dots : Δ v_{6} = 2.46350 \dots

f (v_{5}) = 0.47562 \dots^{2} - 0.49625 \dots \cdot 0.47562 \dots + 1.13068 \dots = 1.12087 \dots

f^{^{'}} (v_{5}) = 2 \cdot 0.47562 \dots - 0.49625 \dots = 0.45499 \dots

v_{6} = - 1.98788 \dots

Δ v_{6} = ∣ - 1.98788 \dots - 0.47562 \dots ∣ = 2.46350 \dots

Δ v_{6} = 2.46350 \dots

v_{7} = - 0.63081 \dots : Δ v_{7} = 1.35707 \dots

f (v_{6}) = (- 1.98788 \dots)^{2} - 0.49625 \dots (- 1.98788 \dots) + 1.13068 \dots = 6.06887 \dots

f^{^{'}} (v_{6}) = 2 (- 1.98788 \dots) - 0.49625 \dots = - 4.47202 \dots

v_{7} = - 0.63081 \dots

Δ v_{7} = ∣ - 0.63081 \dots - (- 1.98788 \dots) ∣ = 1.35707 \dots

Δ v_{7} = 1.35707 \dots

v_{8} = 0.41684 \dots : Δ v_{8} = 1.04765 \dots

f (v_{7}) = (- 0.63081 \dots)^{2} - 0.49625 \dots (- 0.63081 \dots) + 1.13068 \dots = 1.84165 \dots

f^{^{'}} (v_{7}) = 2 (- 0.63081 \dots) - 0.49625 \dots = - 1.75787 \dots

v_{8} = 0.41684 \dots

Δ v_{8} = ∣ 0.41684 \dots - (- 0.63081 \dots) ∣ = 1.04765 \dots

Δ v_{8} = 1.04765 \dots

v_{9} = - 2.83587 \dots : Δ v_{9} = 3.25272 \dots

f (v_{8}) = 0.41684 \dots^{2} - 0.49625 \dots \cdot 0.41684 \dots + 1.13068 \dots = 1.09758 \dots

f^{^{'}} (v_{8}) = 2 \cdot 0.41684 \dots - 0.49625 \dots = 0.33743 \dots

v_{9} = - 2.83587 \dots

Δ v_{9} = ∣ - 2.83587 \dots - 0.41684 \dots ∣ = 3.25272 \dots

Δ v_{9} = 3.25272 \dots

无法得出解

解为

v \approx 0.72449 \dots, v \approx - 1.22074 \dots

v \approx 0.72449 \dots, v \approx - 1.22074 \dots

代回

v = u^{2}

，求解

u

解

u^{2} = 0.72449 \dots : u = 0.72449 \dots, u = - 0.72449 \dots

u^{2} = 0.72449 \dots

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 0.72449 \dots, u = - 0.72449 \dots

解

u^{2} = - 1.22074 \dots : u \in R

无解

u^{2} = - 1.22074 \dots

x^{2}

在

x

内不能为负

\in R

u \in R 无解

解为

u = 0.72449 \dots, u = - 0.72449 \dots

u = sin (x)

代回

sin (x) = 0.72449 \dots, sin (x) = - 0.72449 \dots

sin (x) = 0.72449 \dots, sin (x) = - 0.72449 \dots

sin (x) = 0.72449 \dots : x = arcsin (0.72449 \dots) + 2 πn, x = π - arcsin (0.72449 \dots) + 2 πn

sin (x) = 0.72449 \dots

使用反三角函数性质

sin (x) = 0.72449 \dots

sin (x) = 0.72449 \dots

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (0.72449 \dots) + 2 πn, x = π - arcsin (0.72449 \dots) + 2 πn

x = arcsin (0.72449 \dots) + 2 πn, x = π - arcsin (0.72449 \dots) + 2 πn

sin (x) = - 0.72449 \dots : x = arcsin (- 0.72449 \dots) + 2 πn, x = π + arcsin (0.72449 \dots) + 2 πn

sin (x) = - 0.72449 \dots

使用反三角函数性质

sin (x) = - 0.72449 \dots

sin (x) = - 0.72449 \dots

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- 0.72449 \dots) + 2 πn, x = π + arcsin (0.72449 \dots) + 2 πn

x = arcsin (- 0.72449 \dots) + 2 πn, x = π + arcsin (0.72449 \dots) + 2 πn

合并所有解

x = arcsin (0.72449 \dots) + 2 πn, x = π - arcsin (0.72449 \dots) + 2 πn, x = arcsin (- 0.72449 \dots) + 2 πn, x = π + arcsin (0.72449 \dots) + 2 πn

将解代入原方程进行验证

将它们代入

sin^{4} (x) = - cos (x)

检验解是否符合

去除与方程不符的解。

检验

arcsin (0.72449 \dots) + 2 πn

的解:

假

arcsin (0.72449 \dots) + 2 πn

代入

n = 1

arcsin (0.72449 \dots) + 2 π 1

对于

sin^{4} (x) = - cos (x)

代入

x = arcsin (0.72449 \dots) + 2 π 1

sin^{4} (arcsin (0.72449 \dots) + 2 π 1) = - cos (arcsin (0.72449 \dots) + 2 π 1)

整理后得

0.52488 \dots = - 0.52488 \dots

\Rightarrow 假

检验

π - arcsin (0.72449 \dots) + 2 πn

的解:

真

π - arcsin (0.72449 \dots) + 2 πn

代入

n = 1

π - arcsin (0.72449 \dots) + 2 π 1

对于

sin^{4} (x) = - cos (x)

代入

x = π - arcsin (0.72449 \dots) + 2 π 1

sin^{4} (π - arcsin (0.72449 \dots) + 2 π 1) = - cos (π - arcsin (0.72449 \dots) + 2 π 1)

整理后得

0.52488 \dots = 0.52488 \dots

\Rightarrow 真

检验

arcsin (- 0.72449 \dots) + 2 πn

的解:

假

arcsin (- 0.72449 \dots) + 2 πn

代入

n = 1

arcsin (- 0.72449 \dots) + 2 π 1

对于

sin^{4} (x) = - cos (x)

代入

x = arcsin (- 0.72449 \dots) + 2 π 1

sin^{4} (arcsin (- 0.72449 \dots) + 2 π 1) = - cos (arcsin (- 0.72449 \dots) + 2 π 1)

整理后得

0.52488 \dots = - 0.52488 \dots

\Rightarrow 假

检验

π + arcsin (0.72449 \dots) + 2 πn

的解:

真

π + arcsin (0.72449 \dots) + 2 πn

代入

n = 1

π + arcsin (0.72449 \dots) + 2 π 1

对于

sin^{4} (x) = - cos (x)

代入

x = π + arcsin (0.72449 \dots) + 2 π 1

sin^{4} (π + arcsin (0.72449 \dots) + 2 π 1) = - cos (π + arcsin (0.72449 \dots) + 2 π 1)

整理后得

0.52488 \dots = 0.52488 \dots

\Rightarrow 真

x = π - arcsin (0.72449 \dots) + 2 πn, x = π + arcsin (0.72449 \dots) + 2 πn

以小数形式表示解

x = π - 1.01821 \dots + 2 πn, x = π + 1.01821 \dots + 2 πn

作图

流行的例子

cos^3(x)=4cos^3(x)-3cos(x)

cos^{3} (x) = 4 cos^{3} (x) - 3 cos (x)

4sin^4(x)+12cos^2(x)-7=0

4 sin^{4} (x) + 12 cos^{2} (x) - 7 = 0

5 sin (a) = 4

2cos^4(x)+8sin^2(x)=5

2 cos^{4} (x) + 8 sin^{2} (x) = 5

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

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