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 >

2sin^2(θ)+cos(θ)=0,0<= θ<= 2pi

解答

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

解答

θ = 2.46670 \dots, θ = - 2.46670 \dots + 2 π

+1

度数

θ = 141.33171 \dots^{\circ}, θ = 218.66828 \dots^{\circ}

求解步骤

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

使用三角恒等式改写

cos (θ) + 2 sin^{2} (θ)

使用毕达哥拉斯恒等式:

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

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

= cos (θ) + 2 (1 - cos^{2} (θ))

cos (θ) + (1 - cos^{2} (θ)) \cdot 2 = 0

用替代法求解

cos (θ) + (1 - cos^{2} (θ)) \cdot 2 = 0

令

: cos (θ) = u

u + (1 - u^{2}) \cdot 2 = 0

u + (1 - u^{2}) \cdot 2 = 0 : u = - \frac{- 1 + 17}{4}, u = \frac{1 + 17}{4}

u + (1 - u^{2}) \cdot 2 = 0

展开

u + (1 - u^{2}) \cdot 2 : u + 2 - 2 u^{2}

u + (1 - u^{2}) \cdot 2

= u + 2 (1 - u^{2})

乘开

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

= 2 \cdot 1 - 2 u^{2}

数字相乘:

2 \cdot 1 = 2

= 2 - 2 u^{2}

= u + 2 - 2 u^{2}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

1^{2} - 4 (- 2) \cdot 2 = 17

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 2

使用法则

- (- a) = a

= 1 + 4 \cdot 2 \cdot 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 1 + 16

数字相加:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- 1 \pm 17}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 1 + 17}{2 ( - 2 )}, u_{2} = \frac{- 1 - 17}{2 ( - 2 )}

u = \frac{- 1 + 17}{2 ( - 2 )} : - \frac{- 1 + 17}{4}

\frac{- 1 + 17}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 1 + 17}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 1 + 17}{- 4}

使用分式法则:

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

= - \frac{- 1 + 17}{4}

u = \frac{- 1 - 17}{2 ( - 2 )} : \frac{1 + 17}{4}

\frac{- 1 - 17}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 1 - 17}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 1 - 17}{- 4}

使用分式法则:

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

- 1 - 17 = - (1 + 17)

= \frac{1 + 17}{4}

二次方程组的解是:

u = - \frac{- 1 + 17}{4}, u = \frac{1 + 17}{4}

u = cos (θ)

代回

cos (θ) = - \frac{- 1 + 17}{4}, cos (θ) = \frac{1 + 17}{4}

cos (θ) = - \frac{- 1 + 17}{4}, cos (θ) = \frac{1 + 17}{4}

cos (θ) = - \frac{- 1 + 17}{4}, 0 \leq θ \leq 2 π : θ = arccos (- \frac{17 - 1}{4}), θ = - arccos (- \frac{- 1 + 17}{4}) + 2 π

cos (θ) = - \frac{- 1 + 17}{4}, 0 \leq θ \leq 2 π

使用反三角函数性质

cos (θ) = - \frac{- 1 + 17}{4}

cos (θ) = - \frac{- 1 + 17}{4}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- \frac{- 1 + 17}{4}) + 2 πn, θ = - arccos (- \frac{- 1 + 17}{4}) + 2 πn

θ = arccos (- \frac{- 1 + 17}{4}) + 2 πn, θ = - arccos (- \frac{- 1 + 17}{4}) + 2 πn

在

0 \leq θ \leq 2 π

范围内的解

θ = arccos (- \frac{17 - 1}{4}), θ = - arccos (- \frac{- 1 + 17}{4}) + 2 π

cos (θ) = \frac{1 + 17}{4}, 0 \leq θ \leq 2 π :

无解

cos (θ) = \frac{1 + 17}{4}, 0 \leq θ \leq 2 π

- 1 \leq cos (x) \leq 1

无解

合并所有解

θ = arccos (- \frac{17 - 1}{4}), θ = - arccos (- \frac{- 1 + 17}{4}) + 2 π

以小数形式表示解

θ = 2.46670 \dots, θ = - 2.46670 \dots + 2 π

作图

流行的例子

cos(a)-1=2cos(a)

cos (a) - 1 = 2 cos (a)

3cos^2(θ)-sin(θ)=1

3 cos^{2} (θ) - sin (θ) = 1

5cos(2x)=9sin(x)+4

5 cos (2 x) = 9 sin (x) + 4

-sqrt(3)cos(x)-1=cos(2x)

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

-2tan^2(θ)-4tan(θ)+2=3tan(θ)+5

- 2 tan^{2} (θ) - 4 tan (θ) + 2 = 3 tan (θ) + 5