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受欢迎的三角函数 >

2tan^2(x)-5tan(x)+3=0

解答

2 tan^{2} (x) - 5 tan (x) + 3 = 0

解答

x = 0.98279 \dots + πn, x = \frac{π}{4} + πn

+1

度数

x = 56.30993 \dots^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

2 tan^{2} (x) - 5 tan (x) + 3 = 0

用替代法求解

2 tan^{2} (x) - 5 tan (x) + 3 = 0

令

: tan (x) = u

2 u^{2} - 5 u + 3 = 0

2 u^{2} - 5 u + 3 = 0 : u = \frac{3}{2}, u = 1

2 u^{2} - 5 u + 3 = 0

使用求根公式求解

2 u^{2} - 5 u + 3 = 0

二次方程求根公式:

若

a = 2, b = - 5, c = 3

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

(- 5)^{2} - 4 \cdot 2 \cdot 3 = 1

(- 5)^{2} - 4 \cdot 2 \cdot 3

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 2 \cdot 3

数字相乘:

4 \cdot 2 \cdot 3 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

数字相减:

25 - 24 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 2}, u_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 2}

u = \frac{- ( - 5 ) + 1}{2 \cdot 2} : \frac{3}{2}

\frac{- ( - 5 ) + 1}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{5 + 1}{2 \cdot 2}

数字相加:

5 + 1 = 6

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

数字相乘:

2 \cdot 2 = 4

= \frac{6}{4}

约分:

2

= \frac{3}{2}

u = \frac{- ( - 5 ) - 1}{2 \cdot 2} : 1

\frac{- ( - 5 ) - 1}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{5 - 1}{2 \cdot 2}

数字相减:

5 - 1 = 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

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

u = tan (x)

代回

tan (x) = \frac{3}{2}, tan (x) = 1

tan (x) = \frac{3}{2}, tan (x) = 1

tan (x) = \frac{3}{2} : x = arctan (\frac{3}{2}) + πn

tan (x) = \frac{3}{2}

使用反三角函数性质

tan (x) = \frac{3}{2}

tan (x) = \frac{3}{2}

的通解

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

x = arctan (\frac{3}{2}) + πn

x = arctan (\frac{3}{2}) + πn

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

tan (x) = 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}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

合并所有解

x = arctan (\frac{3}{2}) + πn, x = \frac{π}{4} + πn

以小数形式表示解

x = 0.98279 \dots + πn, x = \frac{π}{4} + πn

作图

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