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受欢迎的三角函数 >

8tan^2(θ)+10tan(θ)+10=7

解答

8 tan^{2} (θ) + 10 tan (θ) + 10 = 7

解答

θ = - 0.46364 \dots + πn, θ = - 0.64350 \dots + πn

+1

度数

θ = - 26.56505 \dots^{\circ} + 18 0^{\circ} n, θ = - 36.86989 \dots^{\circ} + 18 0^{\circ} n

求解步骤

8 tan^{2} (θ) + 10 tan (θ) + 10 = 7

用替代法求解

8 tan^{2} (θ) + 10 tan (θ) + 10 = 7

令

: tan (θ) = u

8 u^{2} + 10 u + 10 = 7

8 u^{2} + 10 u + 10 = 7 : u = - \frac{1}{2}, u = - \frac{3}{4}

8 u^{2} + 10 u + 10 = 7

将

7

para o lado esquerdo

8 u^{2} + 10 u + 10 = 7

两边减去

7

8 u^{2} + 10 u + 10 - 7 = 7 - 7

化简

8 u^{2} + 10 u + 3 = 0

8 u^{2} + 10 u + 3 = 0

使用求根公式求解

8 u^{2} + 10 u + 3 = 0

二次方程求根公式:

若

a = 8, b = 10, c = 3

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 8 \cdot 3}{2 \cdot 8}

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 8 \cdot 3}{2 \cdot 8}

1 0^{2} - 4 \cdot 8 \cdot 3 = 2

1 0^{2} - 4 \cdot 8 \cdot 3

数字相乘:

4 \cdot 8 \cdot 3 = 96

= 1 0^{2} - 96

1 0^{2} = 100

= 100 - 96

数字相减:

100 - 96 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

u_{1, 2} = \frac{- 10 \pm 2}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- 10 + 2}{2 \cdot 8}, u_{2} = \frac{- 10 - 2}{2 \cdot 8}

u = \frac{- 10 + 2}{2 \cdot 8} : - \frac{1}{2}

\frac{- 10 + 2}{2 \cdot 8}

数字相加/相减:

- 10 + 2 = - 8

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

数字相乘:

2 \cdot 8 = 16

= \frac{- 8}{16}

使用分式法则:

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

= - \frac{8}{16}

约分:

8

= - \frac{1}{2}

u = \frac{- 10 - 2}{2 \cdot 8} : - \frac{3}{4}

\frac{- 10 - 2}{2 \cdot 8}

数字相减:

- 10 - 2 = - 12

= \frac{- 12}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 12}{16}

使用分式法则:

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

= - \frac{12}{16}

约分:

4

= - \frac{3}{4}

二次方程组的解是:

u = - \frac{1}{2}, u = - \frac{3}{4}

u = tan (θ)

代回

tan (θ) = - \frac{1}{2}, tan (θ) = - \frac{3}{4}

tan (θ) = - \frac{1}{2}, tan (θ) = - \frac{3}{4}

tan (θ) = - \frac{1}{2} : θ = arctan (- \frac{1}{2}) + πn

tan (θ) = - \frac{1}{2}

使用反三角函数性质

tan (θ) = - \frac{1}{2}

tan (θ) = - \frac{1}{2}

的通解

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

θ = arctan (- \frac{1}{2}) + πn

θ = arctan (- \frac{1}{2}) + πn

tan (θ) = - \frac{3}{4} : θ = arctan (- \frac{3}{4}) + πn

tan (θ) = - \frac{3}{4}

使用反三角函数性质

tan (θ) = - \frac{3}{4}

tan (θ) = - \frac{3}{4}

的通解

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

θ = arctan (- \frac{3}{4}) + πn

θ = arctan (- \frac{3}{4}) + πn

合并所有解

θ = arctan (- \frac{1}{2}) + πn, θ = arctan (- \frac{3}{4}) + πn

以小数形式表示解

θ = - 0.46364 \dots + πn, θ = - 0.64350 \dots + πn

作图

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