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受欢迎的三角函数 >

8tan^2(θ)-1=7tan(θ)

解答

8 tan^{2} (θ) - 1 = 7 tan (θ)

解答

θ = \frac{π}{4} + πn, θ = - 0.12435 \dots + πn

+1

度数

θ = 4 5^{\circ} + 18 0^{\circ} n, θ = - 7.12501 \dots^{\circ} + 18 0^{\circ} n

求解步骤

8 tan^{2} (θ) - 1 = 7 tan (θ)

用替代法求解

8 tan^{2} (θ) - 1 = 7 tan (θ)

令

: tan (θ) = u

8 u^{2} - 1 = 7 u

8 u^{2} - 1 = 7 u : u = 1, u = - \frac{1}{8}

8 u^{2} - 1 = 7 u

将

7 u

para o lado esquerdo

8 u^{2} - 1 = 7 u

两边减去

7 u

8 u^{2} - 1 - 7 u = 7 u - 7 u

化简

8 u^{2} - 1 - 7 u = 0

8 u^{2} - 1 - 7 u = 0

改写成标准形式

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

8 u^{2} - 7 u - 1 = 0

使用求根公式求解

8 u^{2} - 7 u - 1 = 0

二次方程求根公式:

若

a = 8, b = - 7, c = - 1

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 8 ( - 1 )}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 8 ( - 1 )}{2 \cdot 8}

(- 7)^{2} - 4 \cdot 8 (- 1) = 9

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

(- 7)^{2} = 7^{2}

= 7^{2} + 4 \cdot 8 \cdot 1

数字相乘:

4 \cdot 8 \cdot 1 = 32

= 7^{2} + 32

7^{2} = 49

= 49 + 32

数字相加:

49 + 32 = 81

= 81

因式分解数字:

81 = 9^{2}

= 9^{2}

使用根式运算法则:

n a^{n} = a

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 7 ) \pm 9}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- ( - 7 ) + 9}{2 \cdot 8}, u_{2} = \frac{- ( - 7 ) - 9}{2 \cdot 8}

u = \frac{- ( - 7 ) + 9}{2 \cdot 8} : 1

\frac{- ( - 7 ) + 9}{2 \cdot 8}

使用法则

- (- a) = a

= \frac{7 + 9}{2 \cdot 8}

数字相加:

7 + 9 = 16

= \frac{16}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{16}{16}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 7 ) - 9}{2 \cdot 8} : - \frac{1}{8}

\frac{- ( - 7 ) - 9}{2 \cdot 8}

使用法则

- (- a) = a

= \frac{7 - 9}{2 \cdot 8}

数字相减:

7 - 9 = - 2

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

数字相乘:

2 \cdot 8 = 16

= \frac{- 2}{16}

使用分式法则:

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

= - \frac{2}{16}

约分:

2

= - \frac{1}{8}

二次方程组的解是:

u = 1, u = - \frac{1}{8}

u = tan (θ)

代回

tan (θ) = 1, tan (θ) = - \frac{1}{8}

tan (θ) = 1, tan (θ) = - \frac{1}{8}

tan (θ) = 1 : θ = \frac{π}{4} + πn

tan (θ) = 1

tan (θ) = 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

tan (θ) = - \frac{1}{8} : θ = arctan (- \frac{1}{8}) + πn

tan (θ) = - \frac{1}{8}

使用反三角函数性质

tan (θ) = - \frac{1}{8}

tan (θ) = - \frac{1}{8}

的通解

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

θ = arctan (- \frac{1}{8}) + πn

θ = arctan (- \frac{1}{8}) + πn

合并所有解

θ = \frac{π}{4} + πn, θ = arctan (- \frac{1}{8}) + πn

以小数形式表示解

θ = \frac{π}{4} + πn, θ = - 0.12435 \dots + πn

作图

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