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受欢迎的三角函数 >

4tan^2(θ)=11tan(θ)+3

解答

4 tan^{2} (θ) = 11 tan (θ) + 3

解答

θ = 1.24904 \dots + πn, θ = - 0.24497 \dots + πn

+1

度数

θ = 71.56505 \dots^{\circ} + 18 0^{\circ} n, θ = - 14.03624 \dots^{\circ} + 18 0^{\circ} n

求解步骤

4 tan^{2} (θ) = 11 tan (θ) + 3

用替代法求解

4 tan^{2} (θ) = 11 tan (θ) + 3

令

: tan (θ) = u

4 u^{2} = 11 u + 3

4 u^{2} = 11 u + 3 : u = 3, u = - \frac{1}{4}

4 u^{2} = 11 u + 3

将

3

para o lado esquerdo

4 u^{2} = 11 u + 3

两边减去

3

4 u^{2} - 3 = 11 u + 3 - 3

化简

4 u^{2} - 3 = 11 u

4 u^{2} - 3 = 11 u

将

11 u

para o lado esquerdo

4 u^{2} - 3 = 11 u

两边减去

11 u

4 u^{2} - 3 - 11 u = 11 u - 11 u

化简

4 u^{2} - 3 - 11 u = 0

4 u^{2} - 3 - 11 u = 0

改写成标准形式

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

4 u^{2} - 11 u - 3 = 0

使用求根公式求解

4 u^{2} - 11 u - 3 = 0

二次方程求根公式:

若

a = 4, b = - 11, c = - 3

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 4 ( - 3 )}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 4 ( - 3 )}{2 \cdot 4}

(- 11)^{2} - 4 \cdot 4 (- 3) = 13

(- 11)^{2} - 4 \cdot 4 (- 3)

使用法则

- (- a) = a

= (- 11)^{2} + 4 \cdot 4 \cdot 3

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 11)^{2} = 1 1^{2}

= 1 1^{2} + 4 \cdot 4 \cdot 3

数字相乘:

4 \cdot 4 \cdot 3 = 48

= 1 1^{2} + 48

1 1^{2} = 121

= 121 + 48

数字相加:

121 + 48 = 169

= 169

因式分解数字:

169 = 1 3^{2}

= 1 3^{2}

使用根式运算法则:

n a^{n} = a

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- ( - 11 ) \pm 13}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- ( - 11 ) + 13}{2 \cdot 4}, u_{2} = \frac{- ( - 11 ) - 13}{2 \cdot 4}

u = \frac{- ( - 11 ) + 13}{2 \cdot 4} : 3

\frac{- ( - 11 ) + 13}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{11 + 13}{2 \cdot 4}

数字相加:

11 + 13 = 24

= \frac{24}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{24}{8}

数字相除:

\frac{24}{8} = 3

= 3

u = \frac{- ( - 11 ) - 13}{2 \cdot 4} : - \frac{1}{4}

\frac{- ( - 11 ) - 13}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{11 - 13}{2 \cdot 4}

数字相减:

11 - 13 = - 2

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 2}{8}

使用分式法则:

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

= - \frac{2}{8}

约分:

2

= - \frac{1}{4}

二次方程组的解是:

u = 3, u = - \frac{1}{4}

u = tan (θ)

代回

tan (θ) = 3, tan (θ) = - \frac{1}{4}

tan (θ) = 3, tan (θ) = - \frac{1}{4}

tan (θ) = 3 : θ = arctan (3) + πn

tan (θ) = 3

使用反三角函数性质

tan (θ) = 3

tan (θ) = 3

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

θ = arctan (3) + πn

θ = arctan (3) + πn

tan (θ) = - \frac{1}{4} : θ = arctan (- \frac{1}{4}) + πn

tan (θ) = - \frac{1}{4}

使用反三角函数性质

tan (θ) = - \frac{1}{4}

tan (θ) = - \frac{1}{4}

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

θ = arctan (- \frac{1}{4}) + πn

θ = arctan (- \frac{1}{4}) + πn

合并所有解

θ = arctan (3) + πn, θ = arctan (- \frac{1}{4}) + πn

以小数形式表示解

θ = 1.24904 \dots + πn, θ = - 0.24497 \dots + πn

作图

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