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受欢迎的三角函数 >

2sec^2(x)=3tan(x)+1

解答

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

解答

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

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n, x = 26.56505 \dots^{\circ} + 18 0^{\circ} n

求解步骤

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

两边减去

3 tan (x) + 1

2 sec^{2} (x) - 3 tan (x) - 1 = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

化简

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

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

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

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

化简

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

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

对同类项分组

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

数字相加/相减:

- 1 + 2 = 1

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

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

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

1 + 2 tan^{2} (x) - 3 tan (x) = 0

用替代法求解

1 + 2 tan^{2} (x) - 3 tan (x) = 0

令

: tan (x) = u

1 + 2 u^{2} - 3 u = 0

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

1 + 2 u^{2} - 3 u = 0

改写成标准形式

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

2 u^{2} - 3 u + 1 = 0

使用求根公式求解

2 u^{2} - 3 u + 1 = 0

二次方程求根公式:

若

a = 2, b = - 3, c = 1

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

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

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

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

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

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

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数字相减:

9 - 8 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

3 + 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

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

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

使用法则

- (- a) = a

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

数字相减:

3 - 1 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

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

u = tan (x)

代回

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

tan (x) = 1, tan (x) = \frac{1}{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) = \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 = \frac{π}{4} + πn, x = arctan (\frac{1}{2}) + πn

以小数形式表示解

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

作图

流行的例子

2sec(t)=tan^2(t)+1

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

sqrt(2)sin(x)-sqrt(2)cos(x)=sqrt(3)

2 sin (x) - 2 cos (x) = 3

8cos(2x)=4sqrt(3)

8 cos (2 x) = 43

cos^2(θ)-sin^2(θ)=1

cos^{2} (θ) - sin^{2} (θ) = 1

2cos^2(x)=1-cos(x)

2 cos^{2} (x) = 1 - cos (x)