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 >

6cos^2(θ)+sin^2(θ)=4

解答

6 cos^{2} (θ) + sin^{2} (θ) = 4

解答

θ = 0.68471 \dots + 2 πn, θ = π - 0.68471 \dots + 2 πn, θ = - 0.68471 \dots + 2 πn, θ = π + 0.68471 \dots + 2 πn

+1

度数

θ = 39.23152 \dots^{\circ} + 36 0^{\circ} n, θ = 140.76847 \dots^{\circ} + 36 0^{\circ} n, θ = - 39.23152 \dots^{\circ} + 36 0^{\circ} n, θ = 219.23152 \dots^{\circ} + 36 0^{\circ} n

求解步骤

6 cos^{2} (θ) + sin^{2} (θ) = 4

两边减去

4

6 cos^{2} (θ) + sin^{2} (θ) - 4 = 0

使用三角恒等式改写

- 4 + sin^{2} (θ) + 6 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)

= - 4 + sin^{2} (θ) + 6 (1 - sin^{2} (θ))

化简

- 4 + sin^{2} (θ) + 6 (1 - sin^{2} (θ)) : - 5 sin^{2} (θ) + 2

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

乘开

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

6 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 6, b = 1, c = sin^{2} (θ)

= 6 \cdot 1 - 6 sin^{2} (θ)

数字相乘:

6 \cdot 1 = 6

= 6 - 6 sin^{2} (θ)

= - 4 + sin^{2} (θ) + 6 - 6 sin^{2} (θ)

化简

- 4 + sin^{2} (θ) + 6 - 6 sin^{2} (θ) : - 5 sin^{2} (θ) + 2

- 4 + sin^{2} (θ) + 6 - 6 sin^{2} (θ)

对同类项分组

= sin^{2} (θ) - 6 sin^{2} (θ) - 4 + 6

同类项相加:

sin^{2} (θ) - 6 sin^{2} (θ) = - 5 sin^{2} (θ)

= - 5 sin^{2} (θ) - 4 + 6

数字相加/相减:

- 4 + 6 = 2

= - 5 sin^{2} (θ) + 2

= - 5 sin^{2} (θ) + 2

= - 5 sin^{2} (θ) + 2

2 - 5 sin^{2} (θ) = 0

用替代法求解

2 - 5 sin^{2} (θ) = 0

令

: sin (θ) = u

2 - 5 u^{2} = 0

2 - 5 u^{2} = 0 : u = \frac{2}{5}, u = - \frac{2}{5}

2 - 5 u^{2} = 0

将

2

到右边

2 - 5 u^{2} = 0

两边减去

2

2 - 5 u^{2} - 2 = 0 - 2

化简

- 5 u^{2} = - 2

- 5 u^{2} = - 2

两边除以

- 5

- 5 u^{2} = - 2

两边除以

- 5

\frac{- 5 u ^{2}}{- 5} = \frac{- 2}{- 5}

化简

u^{2} = \frac{2}{5}

u^{2} = \frac{2}{5}

对于

x^{2} = f (a)

解为

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

u = \frac{2}{5}, u = - \frac{2}{5}

u = sin (θ)

代回

sin (θ) = \frac{2}{5}, sin (θ) = - \frac{2}{5}

sin (θ) = \frac{2}{5}, sin (θ) = - \frac{2}{5}

sin (θ) = \frac{2}{5} : θ = arcsin (\frac{2}{5}) + 2 πn, θ = π - arcsin (\frac{2}{5}) + 2 πn

sin (θ) = \frac{2}{5}

使用反三角函数性质

sin (θ) = \frac{2}{5}

sin (θ) = \frac{2}{5}

的通解

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

θ = arcsin (\frac{2}{5}) + 2 πn, θ = π - arcsin (\frac{2}{5}) + 2 πn

θ = arcsin (\frac{2}{5}) + 2 πn, θ = π - arcsin (\frac{2}{5}) + 2 πn

sin (θ) = - \frac{2}{5} : θ = arcsin (- \frac{2}{5}) + 2 πn, θ = π + arcsin (\frac{2}{5}) + 2 πn

sin (θ) = - \frac{2}{5}

使用反三角函数性质

sin (θ) = - \frac{2}{5}

sin (θ) = - \frac{2}{5}

的通解

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

θ = arcsin (- \frac{2}{5}) + 2 πn, θ = π + arcsin (\frac{2}{5}) + 2 πn

θ = arcsin (- \frac{2}{5}) + 2 πn, θ = π + arcsin (\frac{2}{5}) + 2 πn

合并所有解

θ = arcsin (\frac{2}{5}) + 2 πn, θ = π - arcsin (\frac{2}{5}) + 2 πn, θ = arcsin (- \frac{2}{5}) + 2 πn, θ = π + arcsin (\frac{2}{5}) + 2 πn

以小数形式表示解

θ = 0.68471 \dots + 2 πn, θ = π - 0.68471 \dots + 2 πn, θ = - 0.68471 \dots + 2 πn, θ = π + 0.68471 \dots + 2 πn

作图

流行的例子

15sin(30)=10sin(x)

15 sin (3 0^{\circ}) = 10 sin (x)

sin(0)=-sin(o)cos(0)

sin (0) = - sin (o) cos (0)

sin(x)=3sqrt(2)

sin (x) = 32

88.2sin(x)-12.78=0.1*88.2cos(x)

88.2 sin (x) - 12.78 = 0.1 \cdot 88.2 cos (x)

1+sin(x)=2*cos(x)

1 + sin (x) = 2 \cdot cos (x)