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受欢迎的三角函数 >

sec^2(x)+0.5tan(x)=1

解答

sec^{2} (x) + 0.5 tan (x) = 1

解答

x = πn, x = - 0.46364 \dots + πn

+1

度数

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

求解步骤

sec^{2} (x) + 0.5 tan (x) = 1

两边减去

1

sec^{2} (x) + 0.5 tan (x) - 1 = 0

使用三角恒等式改写

- 1 + sec^{2} (x) + 0.5 tan (x)

使用毕达哥拉斯恒等式:

sec^{2} (x) = tan^{2} (x) + 1

sec^{2} (x) - 1 = tan^{2} (x)

= 0.5 tan (x) + tan^{2} (x)

tan^{2} (x) + 0.5 tan (x) = 0

用替代法求解

tan^{2} (x) + 0.5 tan (x) = 0

令

: tan (x) = u

u^{2} + 0.5 u = 0

u^{2} + 0.5 u = 0 : u = 0, u = - \frac{1}{2}

u^{2} + 0.5 u = 0

在两边乘以

10

u^{2} + 0.5 u = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

u^{2} \cdot 10 + 0.5 u \cdot 10 = 0 \cdot 10

整理后得

10 u^{2} + 5 u = 0

10 u^{2} + 5 u = 0

使用求根公式求解

10 u^{2} + 5 u = 0

二次方程求根公式:

若

a = 10, b = 5, c = 0

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 10 \cdot 0}{2 \cdot 10}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 10 \cdot 0}{2 \cdot 10}

5^{2} - 4 \cdot 10 \cdot 0 = 5

5^{2} - 4 \cdot 10 \cdot 0

使用法则

0 \cdot a = 0

= 5^{2} - 0

5^{2} - 0 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 5

u_{1, 2} = \frac{- 5 \pm 5}{2 \cdot 10}

将解分隔开

u_{1} = \frac{- 5 + 5}{2 \cdot 10}, u_{2} = \frac{- 5 - 5}{2 \cdot 10}

u = \frac{- 5 + 5}{2 \cdot 10} : 0

\frac{- 5 + 5}{2 \cdot 10}

数字相加/相减:

- 5 + 5 = 0

= \frac{0}{2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{0}{20}

使用法则

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

= 0

u = \frac{- 5 - 5}{2 \cdot 10} : - \frac{1}{2}

\frac{- 5 - 5}{2 \cdot 10}

数字相减:

- 5 - 5 = - 10

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

数字相乘:

2 \cdot 10 = 20

= \frac{- 10}{20}

使用分式法则:

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

= - \frac{10}{20}

约分:

10

= - \frac{1}{2}

二次方程组的解是:

u = 0, u = - \frac{1}{2}

u = tan (x)

代回

tan (x) = 0, tan (x) = - \frac{1}{2}

tan (x) = 0, tan (x) = - \frac{1}{2}

tan (x) = 0 : x = πn

tan (x) = 0

tan (x) = 0

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 = 0 + πn

x = 0 + πn

解

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

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

tan (x) = - \frac{1}{2}

使用反三角函数性质

tan (x) = - \frac{1}{2}

tan (x) = - \frac{1}{2}

的通解

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

x = arctan (- \frac{1}{2}) + πn

x = arctan (- \frac{1}{2}) + πn

合并所有解

x = πn, x = arctan (- \frac{1}{2}) + πn

以小数形式表示解

x = πn, x = - 0.46364 \dots + πn

作图

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