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受欢迎的三角函数 >

sec^2(x)+tan(x)=3

解答

sec^{2} (x) + tan (x) = 3

解答

x = \frac{π}{4} + πn, x = - 1.10714 \dots + πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n, x = - 63.43494 \dots^{\circ} + 18 0^{\circ} n

求解步骤

sec^{2} (x) + tan (x) = 3

两边减去

3

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

使用三角恒等式改写

- 3 + sec^{2} (x) + tan (x)

使用毕达哥拉斯恒等式:

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

= - 3 + tan^{2} (x) + 1 + tan (x)

化简

- 3 + tan^{2} (x) + 1 + tan (x) : tan^{2} (x) + tan (x) - 2

- 3 + tan^{2} (x) + 1 + tan (x)

对同类项分组

= tan^{2} (x) + tan (x) - 3 + 1

数字相加/相减:

- 3 + 1 = - 2

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

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

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

用替代法求解

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

令

: tan (x) = u

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

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

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

改写成标准形式

a x^{2} + b x + c = 0

u^{2} + u - 2 = 0

使用求根公式求解

u^{2} + u - 2 = 0

二次方程求根公式:

若

a = 1, b = 1, c = - 2

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

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

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

1^{2} - 4 \cdot 1 \cdot (- 2)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 2)

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 1 + 3}{2 \cdot 1}, u_{2} = \frac{- 1 - 3}{2 \cdot 1}

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

\frac{- 1 + 3}{2 \cdot 1}

数字相加/相减:

- 1 + 3 = 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- 1 - 3}{2 \cdot 1} : - 2

\frac{- 1 - 3}{2 \cdot 1}

数字相减:

- 1 - 3 = - 4

= \frac{- 4}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 4}{2}

使用分式法则:

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

= - \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= - 2

二次方程组的解是:

u = 1, u = - 2

u = tan (x)

代回

tan (x) = 1, tan (x) = - 2

tan (x) = 1, tan (x) = - 2

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

tan (x) = - 2 : x = arctan (- 2) + πn

tan (x) = - 2

使用反三角函数性质

tan (x) = - 2

tan (x) = - 2

的通解

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

x = arctan (- 2) + πn

x = arctan (- 2) + πn

合并所有解

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

以小数形式表示解

x = \frac{π}{4} + πn, x = - 1.10714 \dots + πn

作图

流行的例子

4sin^2(x)-4sin(x)+1=0

4 sin^{2} (x) - 4 sin (x) + 1 = 0

tan (x) = 5

2sin^3(x)=sin(x)

2 sin^{3} (x) = sin (x)

3cos^2(x)+sin^2(x)=2

3 cos^{2} (x) + sin^{2} (x) = 2

sinh (x) = 0