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受欢迎的三角函数 >

8tan^2(θ)+15tan(θ)-15=2tan(θ)-9

解答

8 tan^{2} (θ) + 15 tan (θ) - 15 = 2 tan (θ) - 9

解答

θ = 0.35877 \dots + πn, θ = - 1.10714 \dots + πn

+1

度数

θ = 20.55604 \dots^{\circ} + 18 0^{\circ} n, θ = - 63.43494 \dots^{\circ} + 18 0^{\circ} n

求解步骤

8 tan^{2} (θ) + 15 tan (θ) - 15 = 2 tan (θ) - 9

用替代法求解

8 tan^{2} (θ) + 15 tan (θ) - 15 = 2 tan (θ) - 9

令

: tan (θ) = u

8 u^{2} + 15 u - 15 = 2 u - 9

8 u^{2} + 15 u - 15 = 2 u - 9 : u = \frac{3}{8}, u = - 2

8 u^{2} + 15 u - 15 = 2 u - 9

将

9

para o lado esquerdo

8 u^{2} + 15 u - 15 = 2 u - 9

两边加上

9

8 u^{2} + 15 u - 15 + 9 = 2 u - 9 + 9

化简

8 u^{2} + 15 u - 6 = 2 u

8 u^{2} + 15 u - 6 = 2 u

将

2 u

para o lado esquerdo

8 u^{2} + 15 u - 6 = 2 u

两边减去

2 u

8 u^{2} + 15 u - 6 - 2 u = 2 u - 2 u

化简

8 u^{2} + 13 u - 6 = 0

8 u^{2} + 13 u - 6 = 0

使用求根公式求解

8 u^{2} + 13 u - 6 = 0

二次方程求根公式:

若

a = 8, b = 13, c = - 6

u_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 8 ( - 6 )}{2 \cdot 8}

u_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 8 ( - 6 )}{2 \cdot 8}

1 3^{2} - 4 \cdot 8 (- 6) = 19

1 3^{2} - 4 \cdot 8 (- 6)

使用法则

- (- a) = a

= 1 3^{2} + 4 \cdot 8 \cdot 6

数字相乘:

4 \cdot 8 \cdot 6 = 192

= 1 3^{2} + 192

1 3^{2} = 169

= 169 + 192

数字相加:

169 + 192 = 361

= 361

因式分解数字:

361 = 1 9^{2}

= 1 9^{2}

使用根式运算法则:

n a^{n} = a

1 9^{2} = 19

= 19

u_{1, 2} = \frac{- 13 \pm 19}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- 13 + 19}{2 \cdot 8}, u_{2} = \frac{- 13 - 19}{2 \cdot 8}

u = \frac{- 13 + 19}{2 \cdot 8} : \frac{3}{8}

\frac{- 13 + 19}{2 \cdot 8}

数字相加/相减:

- 13 + 19 = 6

= \frac{6}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{6}{16}

约分:

2

= \frac{3}{8}

u = \frac{- 13 - 19}{2 \cdot 8} : - 2

\frac{- 13 - 19}{2 \cdot 8}

数字相减:

- 13 - 19 = - 32

= \frac{- 32}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 32}{16}

使用分式法则:

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

= - \frac{32}{16}

数字相除:

\frac{32}{16} = 2

= - 2

二次方程组的解是:

u = \frac{3}{8}, u = - 2

u = tan (θ)

代回

tan (θ) = \frac{3}{8}, tan (θ) = - 2

tan (θ) = \frac{3}{8}, tan (θ) = - 2

tan (θ) = \frac{3}{8} : θ = arctan (\frac{3}{8}) + πn

tan (θ) = \frac{3}{8}

使用反三角函数性质

tan (θ) = \frac{3}{8}

tan (θ) = \frac{3}{8}

的通解

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

θ = arctan (\frac{3}{8}) + πn

θ = arctan (\frac{3}{8}) + 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{3}{8}) + πn, θ = arctan (- 2) + πn

以小数形式表示解

θ = 0.35877 \dots + πn, θ = - 1.10714 \dots + πn

作图

流行的例子

6tan(x)=2sqrt(3)

6 tan (x) = 23

tan (- 3 x) = 1

sin^2(x)=cos^2(x/2)

sin^{2} (x) = cos^{2} (\frac{x}{2})

(2 sin (x) - 1) = 0

solvefor x,sin(x)=y

so l v e f or x, sin (x) = y