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受欢迎的三角函数 >

4.45tan^2(θ)-23tan(θ)+16.15=0

解答

4.45 tan^{2} (θ) - 23 tan (θ) + 16.15 = 0

解答

θ = 1.34385 \dots + πn, θ = 0.69752 \dots + πn

+1

度数

θ = 76.99710 \dots^{\circ} + 18 0^{\circ} n, θ = 39.96514 \dots^{\circ} + 18 0^{\circ} n

求解步骤

4.45 tan^{2} (θ) - 23 tan (θ) + 16.15 = 0

用替代法求解

4.45 tan^{2} (θ) - 23 tan (θ) + 16.15 = 0

令

: tan (θ) = u

4.45 u^{2} - 23 u + 16.15 = 0

4.45 u^{2} - 23 u + 16.15 = 0 : u = \frac{230 + 24153}{89}, u = \frac{230 - 24153}{89}

4.45 u^{2} - 23 u + 16.15 = 0

在两边乘以

100

4.45 u^{2} - 23 u + 16.15 = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

2

digits to the right of the decimal point, therefore multiply by

100

4.45 u^{2} \cdot 100 - 23 u \cdot 100 + 16.15 \cdot 100 = 0 \cdot 100

整理后得

445 u^{2} - 2300 u + 1615 = 0

445 u^{2} - 2300 u + 1615 = 0

使用求根公式求解

445 u^{2} - 2300 u + 1615 = 0

二次方程求根公式:

若

a = 445, b = - 2300, c = 1615

u_{1, 2} = \frac{- ( - 2300 ) \pm ( - 2300 ) ^{2} - 4 \cdot 445 \cdot 1615}{2 \cdot 445}

u_{1, 2} = \frac{- ( - 2300 ) \pm ( - 2300 ) ^{2} - 4 \cdot 445 \cdot 1615}{2 \cdot 445}

(- 2300)^{2} - 4 \cdot 445 \cdot 1615 = 1024153

(- 2300)^{2} - 4 \cdot 445 \cdot 1615

使用指数法则:

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

若

n

是偶数

(- 2300)^{2} = 230 0^{2}

= 230 0^{2} - 4 \cdot 445 \cdot 1615

数字相乘:

4 \cdot 445 \cdot 1615 = 2874700

= 230 0^{2} - 2874700

230 0^{2} = 5290000

= 5290000 - 2874700

数字相减:

5290000 - 2874700 = 2415300

= 2415300

2415300

质因数分解:

2^{2} \cdot 3 \cdot 5^{2} \cdot 83 \cdot 97

2415300

= 2^{2} \cdot 5^{2} \cdot 3 \cdot 83 \cdot 97

使用根式运算法则:

n ab = n a n b

= 2^{2} 5^{2} 3 \cdot 83 \cdot 97

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 5^{2} 3 \cdot 83 \cdot 97

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 2 \cdot 5 3 \cdot 83 \cdot 97

整理后得

= 1024153

u_{1, 2} = \frac{- ( - 2300 ) \pm 10 24153}{2 \cdot 445}

将解分隔开

u_{1} = \frac{- ( - 2300 ) + 10 24153}{2 \cdot 445}, u_{2} = \frac{- ( - 2300 ) - 10 24153}{2 \cdot 445}

u = \frac{- ( - 2300 ) + 10 24153}{2 \cdot 445} : \frac{230 + 24153}{89}

\frac{- ( - 2300 ) + 10 24153}{2 \cdot 445}

使用法则

- (- a) = a

= \frac{2300 + 10 24153}{2 \cdot 445}

数字相乘:

2 \cdot 445 = 890

= \frac{2300 + 10 24153}{890}

分解

2300 + 1024153 : 10 (230 + 24153)

2300 + 1024153

改写为

= 10 \cdot 230 + 1024153

因式分解出通项

10

= 10 (230 + 24153)

= \frac{10 ( 230 + 24153 )}{890}

约分:

10

= \frac{230 + 24153}{89}

u = \frac{- ( - 2300 ) - 10 24153}{2 \cdot 445} : \frac{230 - 24153}{89}

\frac{- ( - 2300 ) - 10 24153}{2 \cdot 445}

使用法则

- (- a) = a

= \frac{2300 - 10 24153}{2 \cdot 445}

数字相乘:

2 \cdot 445 = 890

= \frac{2300 - 10 24153}{890}

分解

2300 - 1024153 : 10 (230 - 24153)

2300 - 1024153

改写为

= 10 \cdot 230 - 1024153

因式分解出通项

10

= 10 (230 - 24153)

= \frac{10 ( 230 - 24153 )}{890}

约分:

10

= \frac{230 - 24153}{89}

二次方程组的解是:

u = \frac{230 + 24153}{89}, u = \frac{230 - 24153}{89}

u = tan (θ)

代回

tan (θ) = \frac{230 + 24153}{89}, tan (θ) = \frac{230 - 24153}{89}

tan (θ) = \frac{230 + 24153}{89}, tan (θ) = \frac{230 - 24153}{89}

tan (θ) = \frac{230 + 24153}{89} : θ = arctan (\frac{230 + 24153}{89}) + πn

tan (θ) = \frac{230 + 24153}{89}

使用反三角函数性质

tan (θ) = \frac{230 + 24153}{89}

tan (θ) = \frac{230 + 24153}{89}

的通解

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

θ = arctan (\frac{230 + 24153}{89}) + πn

θ = arctan (\frac{230 + 24153}{89}) + πn

tan (θ) = \frac{230 - 24153}{89} : θ = arctan (\frac{230 - 24153}{89}) + πn

tan (θ) = \frac{230 - 24153}{89}

使用反三角函数性质

tan (θ) = \frac{230 - 24153}{89}

tan (θ) = \frac{230 - 24153}{89}

的通解

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

θ = arctan (\frac{230 - 24153}{89}) + πn

θ = arctan (\frac{230 - 24153}{89}) + πn

合并所有解

θ = arctan (\frac{230 + 24153}{89}) + πn, θ = arctan (\frac{230 - 24153}{89}) + πn

以小数形式表示解

θ = 1.34385 \dots + πn, θ = 0.69752 \dots + πn

作图

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