zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

sin^2(x)+cos(x)+1=0,0<= x<= 2pi

解答

sin^{2} (x) + cos (x) + 1 = 0, 0 \leq x \leq 2 π

解答

x = π

+1

度数

x = 18 0^{\circ}

求解步骤

sin^{2} (x) + cos (x) + 1 = 0, 0 \leq x \leq 2 π

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

1 + cos (x) + 1 - cos^{2} (x)

对同类项分组

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

数字相加:

1 + 1 = 2

= cos (x) - cos^{2} (x) + 2

= cos (x) - cos^{2} (x) + 2

2 + cos (x) - cos^{2} (x) = 0

用替代法求解

2 + cos (x) - cos^{2} (x) = 0

令

: cos (x) = u

2 + u - u^{2} = 0

2 + u - u^{2} = 0 : u = - 1, u = 2

2 + u - u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 1, b = 1, c = 2

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 2}{2 ( - 1 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 2}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 2 = 3

1^{2} - 4 (- 1) \cdot 2

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 2

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- 1 + 3}{2 ( - 1 )}, u_{2} = \frac{- 1 - 3}{2 ( - 1 )}

u = \frac{- 1 + 3}{2 ( - 1 )} : - 1

\frac{- 1 + 3}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 1 + 3}{- 2 \cdot 1}

数字相加/相减:

- 1 + 3 = 2

= \frac{2}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2}{- 2}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2}{2}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- 1 - 3}{2 ( - 1 )} : 2

\frac{- 1 - 3}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 1 - 3}{- 2 \cdot 1}

数字相减:

- 1 - 3 = - 4

= \frac{- 4}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 4}{- 2}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

二次方程组的解是:

u = - 1, u = 2

u = cos (x)

代回

cos (x) = - 1, cos (x) = 2

cos (x) = - 1, cos (x) = 2

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

cos (x) = - 1, 0 \leq x \leq 2 π

cos (x) = - 1

的通解

cos (x)

周期表（周期为

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 = π + 2 πn

x = π + 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = π

cos (x) = 2, 0 \leq x \leq 2 π :

无解

cos (x) = 2, 0 \leq x \leq 2 π

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = π

作图

流行的例子

sin(2x)=(-1)/(sqrt(3))

sin (2 x) = \frac{- 1}{3}

cos(2θ)=1+3sin(θ)

cos (2 θ) = 1 + 3 sin (θ)

cos(2x)-cos(x)=-1,0<= x<2pi

cos (2 x) - cos (x) = - 1, 0 \leq x < 2 π

sin(x)=-1/(sqrt(10))

sin (x) = - \frac{1}{10}

sec^2(θ)+sec(θ)=2,2pi<= θ<= 3pi

sec^{2} (θ) + sec (θ) = 2, 2 π \leq θ \leq 3 π