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受欢迎的三角函数 >

sec(y)+5tan(y)=3cos(y)

解答

sec (y) + 5 tan (y) = 3 cos (y)

解答

y = 0.33983 \dots + 2 πn, y = π - 0.33983 \dots + 2 πn

+1

度数

y = 19.47122 \dots^{\circ} + 36 0^{\circ} n, y = 160.52877 \dots^{\circ} + 36 0^{\circ} n

求解步骤

sec (y) + 5 tan (y) = 3 cos (y)

两边减去

3 cos (y)

sec (y) + 5 tan (y) - 3 cos (y) = 0

用 sin, cos 表示

\frac{1}{cos ( y )} + 5 \cdot \frac{sin ( y )}{cos ( y )} - 3 cos (y) = 0

化简

\frac{1}{cos ( y )} + 5 \cdot \frac{sin ( y )}{cos ( y )} - 3 cos (y) : \frac{1 + 5 sin ( y ) - 3 cos ^{2} ( y )}{cos ( y )}

\frac{1}{cos ( y )} + 5 \cdot \frac{sin ( y )}{cos ( y )} - 3 cos (y)

乘

5 \cdot \frac{sin ( y )}{cos ( y )} : \frac{5 sin ( y )}{cos ( y )}

5 \cdot \frac{sin ( y )}{cos ( y )}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{sin ( y ) \cdot 5}{cos ( y )}

= \frac{1}{cos ( y )} + \frac{5 sin ( y )}{cos ( y )} - 3 cos (y)

合并分式

\frac{1}{cos ( y )} + \frac{5 sin ( y )}{cos ( y )} : \frac{1 + 5 sin ( y )}{cos ( y )}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 + 5 sin ( y )}{cos ( y )}

= \frac{5 sin ( y ) + 1}{cos ( y )} - 3 cos (y)

将项转换为分式:

3 cos (y) = \frac{3 cos ( y ) cos ( y )}{cos ( y )}

= \frac{1 + sin ( y ) \cdot 5}{cos ( y )} - \frac{3 cos ( y ) cos ( y )}{cos ( y )}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 + sin ( y ) \cdot 5 - 3 cos ( y ) cos ( y )}{cos ( y )}

1 + sin (y) \cdot 5 - 3 cos (y) cos (y) = 1 + 5 sin (y) - 3 cos^{2} (y)

1 + sin (y) \cdot 5 - 3 cos (y) cos (y)

3 cos (y) cos (y) = 3 cos^{2} (y)

3 cos (y) cos (y)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cos (y) cos (y) = cos^{1 + 1} (y)

= 3 cos^{1 + 1} (y)

数字相加:

1 + 1 = 2

= 3 cos^{2} (y)

= 1 + 5 sin (y) - 3 cos^{2} (y)

= \frac{1 + 5 sin ( y ) - 3 cos ^{2} ( y )}{cos ( y )}

\frac{1 + 5 sin ( y ) - 3 cos ^{2} ( y )}{cos ( y )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

1 + 5 sin (y) - 3 cos^{2} (y) = 0

两边加上

3 cos^{2} (y)

1 + 5 sin (y) = 3 cos^{2} (y)

两边进行平方

(1 + 5 sin (y))^{2} = (3 cos^{2} (y))^{2}

两边减去

(3 cos^{2} (y))^{2}

(1 + 5 sin (y))^{2} - 9 cos^{4} (y) = 0

分解

(1 + 5 sin (y))^{2} - 9 cos^{4} (y) : (1 + 5 sin (y) + 3 cos^{2} (y)) (1 + 5 sin (y) - 3 cos^{2} (y))

(1 + 5 sin (y))^{2} - 9 cos^{4} (y)

将

(1 + 5 sin (y))^{2} - 9 cos^{4} (y)

改写为

(1 + 5 sin (y))^{2} - (3 cos^{2} (y))^{2}

(1 + 5 sin (y))^{2} - 9 cos^{4} (y)

将

9

改写为

3^{2}

= (1 + 5 sin (y))^{2} - 3^{2} cos^{4} (y)

使用指数法则:

a^{b c} = (a^{b})^{c}

cos^{4} (y) = (cos^{2} (y))^{2}

= (1 + 5 sin (y))^{2} - 3^{2} (cos^{2} (y))^{2}

使用指数法则:

a^{m} b^{m} = (ab)^{m}

3^{2} (cos^{2} (y))^{2} = (3 cos^{2} (y))^{2}

= (1 + 5 sin (y))^{2} - (3 cos^{2} (y))^{2}

= (1 + 5 sin (y))^{2} - (3 cos^{2} (y))^{2}

使用平方差公式:

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

(1 + 5 sin (y))^{2} - (3 cos^{2} (y))^{2} = ((1 + 5 sin (y)) + 3 cos^{2} (y)) ((1 + 5 sin (y)) - 3 cos^{2} (y))

= ((1 + 5 sin (y)) + 3 cos^{2} (y)) ((1 + 5 sin (y)) - 3 cos^{2} (y))

整理后得

= (3 cos^{2} (y) + 5 sin (y) + 1) (5 sin (y) - 3 cos^{2} (y) + 1)

(1 + 5 sin (y) + 3 cos^{2} (y)) (1 + 5 sin (y) - 3 cos^{2} (y)) = 0

分别求解每个部分

1 + 5 sin (y) + 3 cos^{2} (y) = 0 or 1 + 5 sin (y) - 3 cos^{2} (y) = 0

1 + 5 sin (y) + 3 cos^{2} (y) = 0 : y = arcsin (- \frac{- 5 + 73}{6}) + 2 πn, y = π + arcsin (\frac{- 5 + 73}{6}) + 2 πn

1 + 5 sin (y) + 3 cos^{2} (y) = 0

使用三角恒等式改写

1 + 3 cos^{2} (y) + 5 sin (y)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

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

= 1 + 3 (1 - sin^{2} (y)) + 5 sin (y)

化简

1 + 3 (1 - sin^{2} (y)) + 5 sin (y) : 5 sin (y) - 3 sin^{2} (y) + 4

1 + 3 (1 - sin^{2} (y)) + 5 sin (y)

乘开

3 (1 - sin^{2} (y)) : 3 - 3 sin^{2} (y)

3 (1 - sin^{2} (y))

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (y)

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

数字相乘:

3 \cdot 1 = 3

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

= 1 + 3 - 3 sin^{2} (y) + 5 sin (y)

数字相加:

1 + 3 = 4

= 5 sin (y) - 3 sin^{2} (y) + 4

= 5 sin (y) - 3 sin^{2} (y) + 4

4 - 3 sin^{2} (y) + 5 sin (y) = 0

用替代法求解

4 - 3 sin^{2} (y) + 5 sin (y) = 0

令

: sin (y) = u

4 - 3 u^{2} + 5 u = 0

4 - 3 u^{2} + 5 u = 0 : u = - \frac{- 5 + 73}{6}, u = \frac{5 + 73}{6}

4 - 3 u^{2} + 5 u = 0

改写成标准形式

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

- 3 u^{2} + 5 u + 4 = 0

使用求根公式求解

- 3 u^{2} + 5 u + 4 = 0

二次方程求根公式:

若

a = - 3, b = 5, c = 4

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

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

5^{2} - 4 (- 3) \cdot 4 = 73

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 3 \cdot 4 = 48

= 5^{2} + 48

5^{2} = 25

= 25 + 48

数字相加:

25 + 48 = 73

= 73

u_{1, 2} = \frac{- 5 \pm 73}{2 ( - 3 )}

将解分隔开

u_{1} = \frac{- 5 + 73}{2 ( - 3 )}, u_{2} = \frac{- 5 - 73}{2 ( - 3 )}

u = \frac{- 5 + 73}{2 ( - 3 )} : - \frac{- 5 + 73}{6}

\frac{- 5 + 73}{2 ( - 3 )}

去除括号:

(- a) = - a

= \frac{- 5 + 73}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 5 + 73}{- 6}

使用分式法则:

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

= - \frac{- 5 + 73}{6}

u = \frac{- 5 - 73}{2 ( - 3 )} : \frac{5 + 73}{6}

\frac{- 5 - 73}{2 ( - 3 )}

去除括号:

(- a) = - a

= \frac{- 5 - 73}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 5 - 73}{- 6}

使用分式法则:

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

- 5 - 73 = - (5 + 73)

= \frac{5 + 73}{6}

二次方程组的解是:

u = - \frac{- 5 + 73}{6}, u = \frac{5 + 73}{6}

u = sin (y)

代回

sin (y) = - \frac{- 5 + 73}{6}, sin (y) = \frac{5 + 73}{6}

sin (y) = - \frac{- 5 + 73}{6}, sin (y) = \frac{5 + 73}{6}

sin (y) = - \frac{- 5 + 73}{6} : y = arcsin (- \frac{- 5 + 73}{6}) + 2 πn, y = π + arcsin (\frac{- 5 + 73}{6}) + 2 πn

sin (y) = - \frac{- 5 + 73}{6}

使用反三角函数性质

sin (y) = - \frac{- 5 + 73}{6}

sin (y) = - \frac{- 5 + 73}{6}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

y = arcsin (- \frac{- 5 + 73}{6}) + 2 πn, y = π + arcsin (\frac{- 5 + 73}{6}) + 2 πn

y = arcsin (- \frac{- 5 + 73}{6}) + 2 πn, y = π + arcsin (\frac{- 5 + 73}{6}) + 2 πn

sin (y) = \frac{5 + 73}{6} :

无解

sin (y) = \frac{5 + 73}{6}

- 1 \leq sin (x) \leq 1

无解

合并所有解

y = arcsin (- \frac{- 5 + 73}{6}) + 2 πn, y = π + arcsin (\frac{- 5 + 73}{6}) + 2 πn

1 + 5 sin (y) - 3 cos^{2} (y) = 0 : y = arcsin (\frac{1}{3}) + 2 πn, y = π - arcsin (\frac{1}{3}) + 2 πn

1 + 5 sin (y) - 3 cos^{2} (y) = 0

使用三角恒等式改写

1 - 3 cos^{2} (y) + 5 sin (y)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

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

= 1 - 3 (1 - sin^{2} (y)) + 5 sin (y)

化简

1 - 3 (1 - sin^{2} (y)) + 5 sin (y) : 3 sin^{2} (y) + 5 sin (y) - 2

1 - 3 (1 - sin^{2} (y)) + 5 sin (y)

乘开

- 3 (1 - sin^{2} (y)) : - 3 + 3 sin^{2} (y)

- 3 (1 - sin^{2} (y))

使用分配律:

a (b - c) = ab - a c

a = - 3, b = 1, c = sin^{2} (y)

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

使用加减运算法则

- (- a) = a

= - 3 \cdot 1 + 3 sin^{2} (y)

数字相乘:

3 \cdot 1 = 3

= - 3 + 3 sin^{2} (y)

= 1 - 3 + 3 sin^{2} (y) + 5 sin (y)

数字相减:

1 - 3 = - 2

= 3 sin^{2} (y) + 5 sin (y) - 2

= 3 sin^{2} (y) + 5 sin (y) - 2

- 2 + 3 sin^{2} (y) + 5 sin (y) = 0

用替代法求解

- 2 + 3 sin^{2} (y) + 5 sin (y) = 0

令

: sin (y) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

5^{2} - 4 \cdot 3 (- 2) = 7

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 3 \cdot 2 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

数字相加:

25 + 24 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- 5 \pm 7}{2 \cdot 3}

将解分隔开

u_{1} = \frac{- 5 + 7}{2 \cdot 3}, u_{2} = \frac{- 5 - 7}{2 \cdot 3}

u = \frac{- 5 + 7}{2 \cdot 3} : \frac{1}{3}

\frac{- 5 + 7}{2 \cdot 3}

数字相加/相减:

- 5 + 7 = 2

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

数字相乘:

2 \cdot 3 = 6

= \frac{2}{6}

约分:

2

= \frac{1}{3}

u = \frac{- 5 - 7}{2 \cdot 3} : - 2

\frac{- 5 - 7}{2 \cdot 3}

数字相减:

- 5 - 7 = - 12

= \frac{- 12}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 12}{6}

使用分式法则:

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

= - \frac{12}{6}

数字相除:

\frac{12}{6} = 2

= - 2

二次方程组的解是:

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

u = sin (y)

代回

sin (y) = \frac{1}{3}, sin (y) = - 2

sin (y) = \frac{1}{3}, sin (y) = - 2

sin (y) = \frac{1}{3} : y = arcsin (\frac{1}{3}) + 2 πn, y = π - arcsin (\frac{1}{3}) + 2 πn

sin (y) = \frac{1}{3}

使用反三角函数性质

sin (y) = \frac{1}{3}

sin (y) = \frac{1}{3}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

y = arcsin (\frac{1}{3}) + 2 πn, y = π - arcsin (\frac{1}{3}) + 2 πn

y = arcsin (\frac{1}{3}) + 2 πn, y = π - arcsin (\frac{1}{3}) + 2 πn

sin (y) = - 2 :

无解

sin (y) = - 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

y = arcsin (\frac{1}{3}) + 2 πn, y = π - arcsin (\frac{1}{3}) + 2 πn

合并所有解

y = arcsin (- \frac{- 5 + 73}{6}) + 2 πn, y = π + arcsin (\frac{- 5 + 73}{6}) + 2 πn, y = arcsin (\frac{1}{3}) + 2 πn, y = π - arcsin (\frac{1}{3}) + 2 πn

将解代入原方程进行验证

将它们代入

sec (y) + 5 tan (y) = 3 cos (y)

检验解是否符合

去除与方程不符的解。

检验

arcsin (- \frac{- 5 + 73}{6}) + 2 πn

的解:

假

arcsin (- \frac{- 5 + 73}{6}) + 2 πn

代入

n = 1

arcsin (- \frac{- 5 + 73}{6}) + 2 π 1

对于

sec (y) + 5 tan (y) = 3 cos (y)

代入

y = arcsin (- \frac{- 5 + 73}{6}) + 2 π 1

sec (arcsin (- \frac{- 5 + 73}{6}) + 2 π 1) + 5 tan (arcsin (- \frac{- 5 + 73}{6}) + 2 π 1) = 3 cos (arcsin (- \frac{- 5 + 73}{6}) + 2 π 1)

整理后得

- 2.42074 \dots = 2.42074 \dots

\Rightarrow 假

检验

π + arcsin (\frac{- 5 + 73}{6}) + 2 πn

的解:

假

π + arcsin (\frac{- 5 + 73}{6}) + 2 πn

代入

n = 1

π + arcsin (\frac{- 5 + 73}{6}) + 2 π 1

对于

sec (y) + 5 tan (y) = 3 cos (y)

代入

y = π + arcsin (\frac{- 5 + 73}{6}) + 2 π 1

sec (π + arcsin (\frac{- 5 + 73}{6}) + 2 π 1) + 5 tan (π + arcsin (\frac{- 5 + 73}{6}) + 2 π 1) = 3 cos (π + arcsin (\frac{- 5 + 73}{6}) + 2 π 1)

整理后得

2.42074 \dots = - 2.42074 \dots

\Rightarrow 假

检验

arcsin (\frac{1}{3}) + 2 πn

的解:

真

arcsin (\frac{1}{3}) + 2 πn

代入

n = 1

arcsin (\frac{1}{3}) + 2 π 1

对于

sec (y) + 5 tan (y) = 3 cos (y)

代入

y = arcsin (\frac{1}{3}) + 2 π 1

sec (arcsin (\frac{1}{3}) + 2 π 1) + 5 tan (arcsin (\frac{1}{3}) + 2 π 1) = 3 cos (arcsin (\frac{1}{3}) + 2 π 1)

整理后得

2.82842 \dots = 2.82842 \dots

\Rightarrow 真

检验

π - arcsin (\frac{1}{3}) + 2 πn

的解:

真

π - arcsin (\frac{1}{3}) + 2 πn

代入

n = 1

π - arcsin (\frac{1}{3}) + 2 π 1

对于

sec (y) + 5 tan (y) = 3 cos (y)

代入

y = π - arcsin (\frac{1}{3}) + 2 π 1

sec (π - arcsin (\frac{1}{3}) + 2 π 1) + 5 tan (π - arcsin (\frac{1}{3}) + 2 π 1) = 3 cos (π - arcsin (\frac{1}{3}) + 2 π 1)

整理后得

- 2.82842 \dots = - 2.82842 \dots

\Rightarrow 真

y = arcsin (\frac{1}{3}) + 2 πn, y = π - arcsin (\frac{1}{3}) + 2 πn

以小数形式表示解

y = 0.33983 \dots + 2 πn, y = π - 0.33983 \dots + 2 πn

作图

流行的例子

- 2 sinh (t - 2) = 0

cos(x)sin(x)+2cos^2(x)=0

cos (x) sin (x) + 2 cos^{2} (x) = 0

tan (x) = \frac{130}{45}

csc(pi/(42)x)=1

csc (\frac{π}{42} x) = 1

arctan(4-2x)=arctan(2x)

arctan (4 - 2 x) = arctan (2 x)