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受欢迎的三角函数 >

4tan^2(θ)+16tan(θ)=7tan(θ)-2

解答

4 tan^{2} (θ) + 16 tan (θ) = 7 tan (θ) - 2

解答

θ = - 0.24497 \dots + πn, θ = - 1.10714 \dots + πn

+1

度数

θ = - 14.03624 \dots^{\circ} + 18 0^{\circ} n, θ = - 63.43494 \dots^{\circ} + 18 0^{\circ} n

求解步骤

4 tan^{2} (θ) + 16 tan (θ) = 7 tan (θ) - 2

用替代法求解

4 tan^{2} (θ) + 16 tan (θ) = 7 tan (θ) - 2

令

: tan (θ) = u

4 u^{2} + 16 u = 7 u - 2

4 u^{2} + 16 u = 7 u - 2 : u = - \frac{1}{4}, u = - 2

4 u^{2} + 16 u = 7 u - 2

将

2

para o lado esquerdo

4 u^{2} + 16 u = 7 u - 2

两边加上

2

4 u^{2} + 16 u + 2 = 7 u - 2 + 2

化简

4 u^{2} + 16 u + 2 = 7 u

4 u^{2} + 16 u + 2 = 7 u

将

7 u

para o lado esquerdo

4 u^{2} + 16 u + 2 = 7 u

两边减去

7 u

4 u^{2} + 16 u + 2 - 7 u = 7 u - 7 u

化简

4 u^{2} + 9 u + 2 = 0

4 u^{2} + 9 u + 2 = 0

使用求根公式求解

4 u^{2} + 9 u + 2 = 0

二次方程求根公式:

若

a = 4, b = 9, c = 2

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

9^{2} - 4 \cdot 4 \cdot 2 = 7

9^{2} - 4 \cdot 4 \cdot 2

数字相乘:

4 \cdot 4 \cdot 2 = 32

= 9^{2} - 32

9^{2} = 81

= 81 - 32

数字相减:

81 - 32 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- 9 \pm 7}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- 9 + 7}{2 \cdot 4}, u_{2} = \frac{- 9 - 7}{2 \cdot 4}

u = \frac{- 9 + 7}{2 \cdot 4} : - \frac{1}{4}

\frac{- 9 + 7}{2 \cdot 4}

数字相加/相减:

- 9 + 7 = - 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}

u = \frac{- 9 - 7}{2 \cdot 4} : - 2

\frac{- 9 - 7}{2 \cdot 4}

数字相减:

- 9 - 7 = - 16

= \frac{- 16}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 16}{8}

使用分式法则:

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

= - \frac{16}{8}

数字相除:

\frac{16}{8} = 2

= - 2

二次方程组的解是:

u = - \frac{1}{4}, u = - 2

u = tan (θ)

代回

tan (θ) = - \frac{1}{4}, tan (θ) = - 2

tan (θ) = - \frac{1}{4}, tan (θ) = - 2

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

tan (θ) = - 2 : θ = arctan (- 2) + πn

tan (θ) = - 2

使用反三角函数性质

tan (θ) = - 2

tan (θ) = - 2

的通解

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

θ = arctan (- 2) + πn

θ = arctan (- 2) + πn

合并所有解

θ = arctan (- \frac{1}{4}) + πn, θ = arctan (- 2) + πn

以小数形式表示解

θ = - 0.24497 \dots + πn, θ = - 1.10714 \dots + πn

作图

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