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受欢迎的三角函数 >

9cos(x)+9sin(x)tan(x)=18

解答

9 cos (x) + 9 sin (x) tan (x) = 18

解答

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

+1

度数

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

求解步骤

9 cos (x) + 9 sin (x) tan (x) = 18

两边减去

18

9 cos (x) + 9 sin (x) tan (x) - 18 = 0

使用三角恒等式改写

- 18 + 9 cos (x) + 9 sin (x) tan (x)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - 18 + 9 cos (x) + 9 sin (x) \frac{sin ( x )}{cos ( x )}

9 sin (x) \frac{sin ( x )}{cos ( x )} = \frac{9 sin ^{2} ( x )}{cos ( x )}

9 sin (x) \frac{sin ( x )}{cos ( x )}

分式相乘:

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

= \frac{sin ( x ) \cdot 9 sin ( x )}{cos ( x )}

sin (x) \cdot 9 sin (x) = 9 sin^{2} (x)

sin (x) \cdot 9 sin (x)

使用指数法则:

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

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

= 9 sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= 9 sin^{2} (x)

= \frac{9 sin ^{2} ( x )}{cos ( x )}

= - 18 + 9 cos (x) + \frac{9 sin ^{2} ( x )}{cos ( x )}

使用毕达哥拉斯恒等式:

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

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

= - 18 + \frac{9 ( 1 - cos ^{2} ( x ) )}{cos ( x )} + 9 cos (x)

合并分式

\frac{9 ( - cos ^{2} ( x ) + 1 )}{cos ( x )} + 9 cos (x) : \frac{9}{cos ( x )}

\frac{9 ( - cos ^{2} ( x ) + 1 )}{cos ( x )} + 9 cos (x)

将项转换为分式:

9 cos (x) = \frac{9 cos ( x ) cos ( x )}{cos ( x )}

= \frac{9 ( 1 - cos ^{2} ( x ) )}{cos ( x )} + \frac{9 cos ( x ) cos ( x )}{cos ( x )}

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{9 ( 1 - cos ^{2} ( x ) ) + 9 cos ( x ) cos ( x )}{cos ( x )}

9 (1 - cos^{2} (x)) + 9 cos (x) cos (x) = 9 (1 - cos^{2} (x)) + 9 cos^{2} (x)

9 (1 - cos^{2} (x)) + 9 cos (x) cos (x)

9 cos (x) cos (x) = 9 cos^{2} (x)

9 cos (x) cos (x)

使用指数法则:

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

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

= 9 cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= 9 cos^{2} (x)

= 9 (- cos^{2} (x) + 1) + 9 cos^{2} (x)

= \frac{9 ( - cos ^{2} ( x ) + 1 ) + 9 cos ^{2} ( x )}{cos ( x )}

乘开

9 (1 - cos^{2} (x)) + 9 cos^{2} (x) : 9

9 (1 - cos^{2} (x)) + 9 cos^{2} (x)

乘开

9 (1 - cos^{2} (x)) : 9 - 9 cos^{2} (x)

9 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 9, b = 1, c = cos^{2} (x)

= 9 \cdot 1 - 9 cos^{2} (x)

数字相乘:

9 \cdot 1 = 9

= 9 - 9 cos^{2} (x)

= 9 - 9 cos^{2} (x) + 9 cos^{2} (x)

同类项相加:

- 9 cos^{2} (x) + 9 cos^{2} (x) = 0

= 9

= \frac{9}{cos ( x )}

= \frac{9}{cos ( x )} - 18

- 18 + \frac{9}{cos ( x )} = 0

- 18 + \frac{9}{cos ( x )} = 0

在两边乘以

cos (x)

- 18 + \frac{9}{cos ( x )} = 0

在两边乘以

cos (x)

- 18 cos (x) + \frac{9}{cos ( x )} cos (x) = 0 \cdot cos (x)

化简

- 18 cos (x) + \frac{9}{cos ( x )} cos (x) = 0 \cdot cos (x)

化简

\frac{9}{cos ( x )} cos (x) : 9

\frac{9}{cos ( x )} cos (x)

分式相乘:

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

= \frac{9 cos ( x )}{cos ( x )}

约分:

cos (x)

= 9

化简

0 \cdot cos (x) : 0

0 \cdot cos (x)

使用法则

0 \cdot a = 0

= 0

- 18 cos (x) + 9 = 0

- 18 cos (x) + 9 = 0

- 18 cos (x) + 9 = 0

将

9

到右边

- 18 cos (x) + 9 = 0

两边减去

9

- 18 cos (x) + 9 - 9 = 0 - 9

化简

- 18 cos (x) = - 9

- 18 cos (x) = - 9

两边除以

- 18

- 18 cos (x) = - 9

两边除以

- 18

\frac{- 18 cos ( x )}{- 18} = \frac{- 9}{- 18}

化简

\frac{- 18 cos ( x )}{- 18} = \frac{- 9}{- 18}

化简

\frac{- 18 cos ( x )}{- 18} : cos (x)

\frac{- 18 cos ( x )}{- 18}

使用分式法则:

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

= \frac{18 cos ( x )}{18}

数字相除:

\frac{18}{18} = 1

= cos (x)

化简

\frac{- 9}{- 18} : \frac{1}{2}

\frac{- 9}{- 18}

使用分式法则:

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

= \frac{9}{18}

约分:

9

= \frac{1}{2}

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}

验证解

找到无定义的点（奇点）:

cos (x) = 0

取

- 18 + \frac{9}{cos ( x )}

的分母，令其等于零

cos (x) = 0

以下点无定义

cos (x) = 0

将不在定义域的点与解相综合:

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}

的通解

cos (x)

周期表（周期为

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

作图

流行的例子

cot (2 x) = 1

5 tan^{2} (x) - 15 = 0

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

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

sin (θ) = - \frac{2}{5}

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

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