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 >

9sin^2(θ)-1=0

解答

9 sin^{2} (θ) - 1 = 0

解答

θ = 0.33983 \dots + 2 πn, θ = π - 0.33983 \dots + 2 πn, θ = - 0.33983 \dots + 2 πn, θ = π + 0.33983 \dots + 2 πn

+1

度数

θ = 19.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 160.52877 \dots^{\circ} + 36 0^{\circ} n, θ = - 19.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 199.47122 \dots^{\circ} + 36 0^{\circ} n

求解步骤

9 sin^{2} (θ) - 1 = 0

用替代法求解

9 sin^{2} (θ) - 1 = 0

令

: sin (θ) = u

9 u^{2} - 1 = 0

9 u^{2} - 1 = 0 : u = \frac{1}{3}, u = - \frac{1}{3}

9 u^{2} - 1 = 0

将

1

到右边

9 u^{2} - 1 = 0

两边加上

1

9 u^{2} - 1 + 1 = 0 + 1

化简

9 u^{2} = 1

9 u^{2} = 1

两边除以

9

9 u^{2} = 1

两边除以

9

\frac{9 u ^{2}}{9} = \frac{1}{9}

化简

u^{2} = \frac{1}{9}

u^{2} = \frac{1}{9}

对于

x^{2} = f (a)

解为

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

u = \frac{1}{9}, u = - \frac{1}{9}

\frac{1}{9} = \frac{1}{3}

\frac{1}{9}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{1}{9}

9 = 3

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

= \frac{1}{3}

使用法则

1 = 1

= \frac{1}{3}

- \frac{1}{9} = - \frac{1}{3}

- \frac{1}{9}

化简

\frac{1}{9} : \frac{1}{3}

\frac{1}{9}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{1}{9}

9 = 3

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

= \frac{1}{3}

= - \frac{1}{3}

使用法则

1 = 1

= - \frac{1}{3}

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

u = sin (θ)

代回

sin (θ) = \frac{1}{3}, sin (θ) = - \frac{1}{3}

sin (θ) = \frac{1}{3}, sin (θ) = - \frac{1}{3}

sin (θ) = \frac{1}{3} : θ = arcsin (\frac{1}{3}) + 2 πn, θ = π - arcsin (\frac{1}{3}) + 2 πn

sin (θ) = \frac{1}{3}

使用反三角函数性质

sin (θ) = \frac{1}{3}

sin (θ) = \frac{1}{3}

的通解

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

θ = arcsin (\frac{1}{3}) + 2 πn, θ = π - arcsin (\frac{1}{3}) + 2 πn

θ = arcsin (\frac{1}{3}) + 2 πn, θ = π - arcsin (\frac{1}{3}) + 2 πn

sin (θ) = - \frac{1}{3} : θ = arcsin (- \frac{1}{3}) + 2 πn, θ = π + arcsin (\frac{1}{3}) + 2 πn

sin (θ) = - \frac{1}{3}

使用反三角函数性质

sin (θ) = - \frac{1}{3}

sin (θ) = - \frac{1}{3}

的通解

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

θ = arcsin (- \frac{1}{3}) + 2 πn, θ = π + arcsin (\frac{1}{3}) + 2 πn

θ = arcsin (- \frac{1}{3}) + 2 πn, θ = π + arcsin (\frac{1}{3}) + 2 πn

合并所有解

θ = arcsin (\frac{1}{3}) + 2 πn, θ = π - arcsin (\frac{1}{3}) + 2 πn, θ = arcsin (- \frac{1}{3}) + 2 πn, θ = π + arcsin (\frac{1}{3}) + 2 πn

以小数形式表示解

θ = 0.33983 \dots + 2 πn, θ = π - 0.33983 \dots + 2 πn, θ = - 0.33983 \dots + 2 πn, θ = π + 0.33983 \dots + 2 πn

作图

流行的例子

sinh (x) = 3

cos(2θ-pi/2)=1,0<= θ<2pi

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

sin (2 x) = - 0.5

cos (x + \frac{π}{4}) = 0

sin^2(θ)cos(θ)+cos^3(θ)=-1/2

sin^{2} (θ) cos (θ) + cos^{3} (θ) = - \frac{1}{2}