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受欢迎的三角函数 >

7tan^2(θ)-4tan(θ)=0,-90<= θ<= 90

解答

7 tan^{2} (θ) - 4 tan (θ) = 0, - 9 0^{\circ} \leq θ \leq 9 0^{\circ}

解答

θ = 0.51914 \dots, θ = 0

求解步骤

7 tan^{2} (θ) - 4 tan (θ) = 0, - 9 0^{\circ} \leq θ \leq 9 0^{\circ}

用替代法求解

7 tan^{2} (θ) - 4 tan (θ) = 0

令

: tan (θ) = u

7 u^{2} - 4 u = 0

7 u^{2} - 4 u = 0 : u = \frac{4}{7}, u = 0

7 u^{2} - 4 u = 0

使用求根公式求解

7 u^{2} - 4 u = 0

二次方程求根公式:

若

a = 7, b = - 4, c = 0

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 7 \cdot 0}{2 \cdot 7}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 7 \cdot 0}{2 \cdot 7}

(- 4)^{2} - 4 \cdot 7 \cdot 0 = 4

(- 4)^{2} - 4 \cdot 7 \cdot 0

使用指数法则:

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

若

n

是偶数

(- 4)^{2} = 4^{2}

= 4^{2} - 4 \cdot 7 \cdot 0

使用法则

0 \cdot a = 0

= 4^{2} - 0

4^{2} - 0 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 4

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

将解分隔开

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

u = \frac{- ( - 4 ) + 4}{2 \cdot 7} : \frac{4}{7}

\frac{- ( - 4 ) + 4}{2 \cdot 7}

使用法则

- (- a) = a

= \frac{4 + 4}{2 \cdot 7}

数字相加:

4 + 4 = 8

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

数字相乘:

2 \cdot 7 = 14

= \frac{8}{14}

约分:

2

= \frac{4}{7}

u = \frac{- ( - 4 ) - 4}{2 \cdot 7} : 0

\frac{- ( - 4 ) - 4}{2 \cdot 7}

使用法则

- (- a) = a

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

数字相减:

4 - 4 = 0

= \frac{0}{2 \cdot 7}

数字相乘:

2 \cdot 7 = 14

= \frac{0}{14}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

二次方程组的解是:

u = \frac{4}{7}, u = 0

u = tan (θ)

代回

tan (θ) = \frac{4}{7}, tan (θ) = 0

tan (θ) = \frac{4}{7}, tan (θ) = 0

tan (θ) = \frac{4}{7}, - 9 0^{\circ} \leq θ \leq 9 0^{\circ} : θ = arctan (\frac{4}{7})

tan (θ) = \frac{4}{7}, - 9 0^{\circ} \leq θ \leq 9 0^{\circ}

使用反三角函数性质

tan (θ) = \frac{4}{7}

tan (θ) = \frac{4}{7}

的通解

tan (x) = a \Rightarrow x = arctan (a) + 18 0^{\circ} n

θ = arctan (\frac{4}{7}) + 18 0^{\circ} n

θ = arctan (\frac{4}{7}) + 18 0^{\circ} n

在

- 9 0^{\circ} \leq θ \leq 9 0^{\circ}

范围内的解

θ = arctan (\frac{4}{7})

tan (θ) = 0, - 9 0^{\circ} \leq θ \leq 9 0^{\circ} : θ = 0

tan (θ) = 0, - 9 0^{\circ} \leq θ \leq 9 0^{\circ}

tan (θ) = 0

的通解

tan (x)

周期表（周期为

18 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = 0 + 18 0^{\circ} n

θ = 0 + 18 0^{\circ} n

解

θ = 0 + 18 0^{\circ} n : θ = 18 0^{\circ} n

θ = 0 + 18 0^{\circ} n

0 + 18 0^{\circ} n = 18 0^{\circ} n

θ = 18 0^{\circ} n

θ = 18 0^{\circ} n

在

- 9 0^{\circ} \leq θ \leq 9 0^{\circ}

范围内的解

θ = 0

合并所有解

θ = arctan (\frac{4}{7}), θ = 0

以小数形式表示解

θ = 0.51914 \dots, θ = 0

作图

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