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 >

7cos^2(θ)+6sin(θ)-10=-4

解答

7 cos^{2} (θ) + 6 sin (θ) - 10 = - 4

解答

θ = - 0.14334 \dots + 2 πn, θ = π + 0.14334 \dots + 2 πn, θ = \frac{π}{2} + 2 πn

+1

度数

θ = - 8.21321 \dots^{\circ} + 36 0^{\circ} n, θ = 188.21321 \dots^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

7 cos^{2} (θ) + 6 sin (θ) - 10 = - 4

两边减去

- 4

7 cos^{2} (θ) + 6 sin (θ) - 6 = 0

使用三角恒等式改写

- 6 + 6 sin (θ) + 7 cos^{2} (θ)

使用毕达哥拉斯恒等式:

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

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

= - 6 + 6 sin (θ) + 7 (1 - sin^{2} (θ))

化简

- 6 + 6 sin (θ) + 7 (1 - sin^{2} (θ)) : 6 sin (θ) - 7 sin^{2} (θ) + 1

- 6 + 6 sin (θ) + 7 (1 - sin^{2} (θ))

乘开

7 (1 - sin^{2} (θ)) : 7 - 7 sin^{2} (θ)

7 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 7, b = 1, c = sin^{2} (θ)

= 7 \cdot 1 - 7 sin^{2} (θ)

数字相乘:

7 \cdot 1 = 7

= 7 - 7 sin^{2} (θ)

= - 6 + 6 sin (θ) + 7 - 7 sin^{2} (θ)

化简

- 6 + 6 sin (θ) + 7 - 7 sin^{2} (θ) : 6 sin (θ) - 7 sin^{2} (θ) + 1

- 6 + 6 sin (θ) + 7 - 7 sin^{2} (θ)

对同类项分组

= 6 sin (θ) - 7 sin^{2} (θ) - 6 + 7

数字相加/相减:

- 6 + 7 = 1

= 6 sin (θ) - 7 sin^{2} (θ) + 1

= 6 sin (θ) - 7 sin^{2} (θ) + 1

= 6 sin (θ) - 7 sin^{2} (θ) + 1

1 + 6 sin (θ) - 7 sin^{2} (θ) = 0

用替代法求解

1 + 6 sin (θ) - 7 sin^{2} (θ) = 0

令

: sin (θ) = u

1 + 6 u - 7 u^{2} = 0

1 + 6 u - 7 u^{2} = 0 : u = - \frac{1}{7}, u = 1

1 + 6 u - 7 u^{2} = 0

改写成标准形式

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

- 7 u^{2} + 6 u + 1 = 0

使用求根公式求解

- 7 u^{2} + 6 u + 1 = 0

二次方程求根公式:

若

a = - 7, b = 6, c = 1

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 7 ) \cdot 1}{2 ( - 7 )}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 7 ) \cdot 1}{2 ( - 7 )}

6^{2} - 4 (- 7) \cdot 1 = 8

6^{2} - 4 (- 7) \cdot 1

使用法则

- (- a) = a

= 6^{2} + 4 \cdot 7 \cdot 1

数字相乘:

4 \cdot 7 \cdot 1 = 28

= 6^{2} + 28

6^{2} = 36

= 36 + 28

数字相加:

36 + 28 = 64

= 64

因式分解数字:

64 = 8^{2}

= 8^{2}

使用根式运算法则:

n a^{n} = a

8^{2} = 8

= 8

u_{1, 2} = \frac{- 6 \pm 8}{2 ( - 7 )}

将解分隔开

u_{1} = \frac{- 6 + 8}{2 ( - 7 )}, u_{2} = \frac{- 6 - 8}{2 ( - 7 )}

u = \frac{- 6 + 8}{2 ( - 7 )} : - \frac{1}{7}

\frac{- 6 + 8}{2 ( - 7 )}

去除括号:

(- a) = - a

= \frac{- 6 + 8}{- 2 \cdot 7}

数字相加/相减:

- 6 + 8 = 2

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

数字相乘:

2 \cdot 7 = 14

= \frac{2}{- 14}

使用分式法则:

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

= - \frac{2}{14}

约分:

2

= - \frac{1}{7}

u = \frac{- 6 - 8}{2 ( - 7 )} : 1

\frac{- 6 - 8}{2 ( - 7 )}

去除括号:

(- a) = - a

= \frac{- 6 - 8}{- 2 \cdot 7}

数字相减:

- 6 - 8 = - 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{1}{7}, u = 1

u = sin (θ)

代回

sin (θ) = - \frac{1}{7}, sin (θ) = 1

sin (θ) = - \frac{1}{7}, sin (θ) = 1

sin (θ) = - \frac{1}{7} : θ = arcsin (- \frac{1}{7}) + 2 πn, θ = π + arcsin (\frac{1}{7}) + 2 πn

sin (θ) = - \frac{1}{7}

使用反三角函数性质

sin (θ) = - \frac{1}{7}

sin (θ) = - \frac{1}{7}

的通解

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

θ = arcsin (- \frac{1}{7}) + 2 πn, θ = π + arcsin (\frac{1}{7}) + 2 πn

θ = arcsin (- \frac{1}{7}) + 2 πn, θ = π + arcsin (\frac{1}{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{1}{7}) + 2 πn, θ = π + arcsin (\frac{1}{7}) + 2 πn, θ = \frac{π}{2} + 2 πn

以小数形式表示解

θ = - 0.14334 \dots + 2 πn, θ = π + 0.14334 \dots + 2 πn, θ = \frac{π}{2} + 2 πn

作图

流行的例子

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

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

0=-2sin(x)+cos(x)

0 = - 2 sin (x) + cos (x)

tan (θ) = \frac{7}{6}

2 sin (x) + 5 = 6

sin(2θ)-cos(2θ)=1

sin (2 θ) - cos (2 θ) = 1