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受欢迎的三角函数 >

2sec^2(x)=3-tan(x),0<= x<= 2pi

解答

2 sec^{2} (x) = 3 - tan (x), 0 \leq x \leq 2 π

解答

x = 0.46364 \dots, x = 0.46364 \dots + π, x = \frac{3 π}{4}, x = \frac{7 π}{4}

+1

度数

x = 26.56505 \dots^{\circ}, x = 206.56505 \dots^{\circ}, x = 13 5^{\circ}, x = 31 5^{\circ}

求解步骤

2 sec^{2} (x) = 3 - tan (x), 0 \leq x \leq 2 π

两边减去

3 - tan (x)

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

化简

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

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

乘开

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

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

使用分配律:

a (b + c) = ab + a c

a = 2, b = tan^{2} (x), c = 1

= 2 tan^{2} (x) + 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

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

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

化简

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

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

对同类项分组

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

数字相加/相减:

- 3 + 2 = - 1

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

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

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

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

用替代法求解

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

令

: tan (x) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 1)

使用法则

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 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 2}

将解分隔开

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

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

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

数字相加/相减:

- 1 + 3 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

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

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

数字相减:

- 1 - 3 = - 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 4}{4}

使用分式法则:

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

= - \frac{4}{4}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

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

u = tan (x)

代回

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

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

tan (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = arctan (\frac{1}{2}), x = arctan (\frac{1}{2}) + π

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

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

在

0 \leq x \leq 2 π

范围内的解

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

tan (x) = - 1, 0 \leq x \leq 2 π : x = \frac{3 π}{4}, x = \frac{7 π}{4}

tan (x) = - 1, 0 \leq x \leq 2 π

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{3 π}{4} + πn

x = \frac{3 π}{4} + πn

在

0 \leq x \leq 2 π

范围内的解

x = \frac{3 π}{4}, x = \frac{7 π}{4}

合并所有解

x = arctan (\frac{1}{2}), x = arctan (\frac{1}{2}) + π, x = \frac{3 π}{4}, x = \frac{7 π}{4}

以小数形式表示解

x = 0.46364 \dots, x = 0.46364 \dots + π, x = \frac{3 π}{4}, x = \frac{7 π}{4}

作图

流行的例子

1+cos(2θ)=-3cos(θ)

1 + cos (2 θ) = - 3 cos (θ)

tan (θ) = \frac{6}{0}

3cos(20t)+6sin(20t)=0

3 cos (20 t) + 6 sin (20 t) = 0

tan (θ) = \frac{6}{2}

tan (θ) = \frac{6}{4}