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 >

-7sin^2(θ)+4sin(θ)=-3

解答

- 7 sin^{2} (θ) + 4 sin (θ) = - 3

解答

θ = - 0.44291 \dots + 2 πn, θ = π + 0.44291 \dots + 2 πn, θ = \frac{π}{2} + 2 πn

+1

度数

θ = - 25.37693 \dots^{\circ} + 36 0^{\circ} n, θ = 205.37693 \dots^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

- 7 sin^{2} (θ) + 4 sin (θ) = - 3

用替代法求解

- 7 sin^{2} (θ) + 4 sin (θ) = - 3

令

: sin (θ) = u

- 7 u^{2} + 4 u = - 3

- 7 u^{2} + 4 u = - 3 : u = - \frac{3}{7}, u = 1

- 7 u^{2} + 4 u = - 3

将

3

para o lado esquerdo

- 7 u^{2} + 4 u = - 3

两边加上

3

- 7 u^{2} + 4 u + 3 = - 3 + 3

化简

- 7 u^{2} + 4 u + 3 = 0

- 7 u^{2} + 4 u + 3 = 0

使用求根公式求解

- 7 u^{2} + 4 u + 3 = 0

二次方程求根公式:

若

a = - 7, b = 4, c = 3

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 7 ) \cdot 3}{2 ( - 7 )}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 7 ) \cdot 3}{2 ( - 7 )}

4^{2} - 4 (- 7) \cdot 3 = 10

4^{2} - 4 (- 7) \cdot 3

使用法则

- (- a) = a

= 4^{2} + 4 \cdot 7 \cdot 3

数字相乘:

4 \cdot 7 \cdot 3 = 84

= 4^{2} + 84

4^{2} = 16

= 16 + 84

数字相加:

16 + 84 = 100

= 100

因式分解数字:

100 = 1 0^{2}

= 1 0^{2}

使用根式运算法则:

n a^{n} = a

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- 4 \pm 10}{2 ( - 7 )}

将解分隔开

u_{1} = \frac{- 4 + 10}{2 ( - 7 )}, u_{2} = \frac{- 4 - 10}{2 ( - 7 )}

u = \frac{- 4 + 10}{2 ( - 7 )} : - \frac{3}{7}

\frac{- 4 + 10}{2 ( - 7 )}

去除括号:

(- a) = - a

= \frac{- 4 + 10}{- 2 \cdot 7}

数字相加/相减:

- 4 + 10 = 6

= \frac{6}{- 2 \cdot 7}

数字相乘:

2 \cdot 7 = 14

= \frac{6}{- 14}

使用分式法则:

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

= - \frac{6}{14}

约分:

2

= - \frac{3}{7}

u = \frac{- 4 - 10}{2 ( - 7 )} : 1

\frac{- 4 - 10}{2 ( - 7 )}

去除括号:

(- a) = - a

= \frac{- 4 - 10}{- 2 \cdot 7}

数字相减:

- 4 - 10 = - 14

= \frac{- 14}{- 2 \cdot 7}

数字相乘:

2 \cdot 7 = 14

= \frac{- 14}{- 14}

使用分式法则:

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

= \frac{14}{14}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = - \frac{3}{7}, u = 1

u = sin (θ)

代回

sin (θ) = - \frac{3}{7}, sin (θ) = 1

sin (θ) = - \frac{3}{7}, sin (θ) = 1

sin (θ) = - \frac{3}{7} : θ = arcsin (- \frac{3}{7}) + 2 πn, θ = π + arcsin (\frac{3}{7}) + 2 πn

sin (θ) = - \frac{3}{7}

使用反三角函数性质

sin (θ) = - \frac{3}{7}

sin (θ) = - \frac{3}{7}

的通解

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

θ = arcsin (- \frac{3}{7}) + 2 πn, θ = π + arcsin (\frac{3}{7}) + 2 πn

θ = arcsin (- \frac{3}{7}) + 2 πn, θ = π + arcsin (\frac{3}{7}) + 2 πn

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = 1

sin (θ) = 1

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

合并所有解

θ = arcsin (- \frac{3}{7}) + 2 πn, θ = π + arcsin (\frac{3}{7}) + 2 πn, θ = \frac{π}{2} + 2 πn

以小数形式表示解

θ = - 0.44291 \dots + 2 πn, θ = π + 0.44291 \dots + 2 πn, θ = \frac{π}{2} + 2 πn

作图

流行的例子

sqrt(3)=sqrt(3)cos(2x)+sin(2x)

3 = 3 cos (2 x) + sin (2 x)

csc(θ)=-(2sqrt(3))/3

csc (θ) = - \frac{2 3}{3}

2sin(x/3)+sqrt(3)=0

2 sin (\frac{x}{3}) + 3 = 0

sin (θ) = 4

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

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