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sin(2x)=tan(x)

解答

sin (2 x) = tan (x)

解答

x = 2 πn, x = π + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

求解步骤

sin (2 x) = tan (x)

两边减去

tan (x)

sin (2 x) - tan (x) = 0

用 sin, cos 表示

sin (2 x) - tan (x)

使用基本三角恒等式:

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

= sin (2 x) - \frac{sin ( x )}{cos ( x )}

化简

sin (2 x) - \frac{sin ( x )}{cos ( x )} : \frac{sin ( 2 x ) cos ( x ) - sin ( x )}{cos ( x )}

sin (2 x) - \frac{sin ( x )}{cos ( x )}

将项转换为分式:

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

= \frac{sin ( 2 x ) cos ( x )}{cos ( x )} - \frac{sin ( 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{sin ( 2 x ) cos ( x ) - sin ( x )}{cos ( x )}

= \frac{sin ( 2 x ) cos ( x ) - sin ( x )}{cos ( x )}

\frac{- sin ( x ) + cos ( x ) sin ( 2 x )}{cos ( x )} = 0

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

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

使用三角恒等式改写

- sin (x) + cos (x) sin (2 x)

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

= - sin (x) + cos (x) \cdot 2 sin (x) cos (x)

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

cos (x) \cdot 2 sin (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)

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

数字相加:

1 + 1 = 2

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

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

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

分解

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

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

因式分解出通项

sin (x)

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

分解

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

2 cos^{2} (x) - 1

将

2 cos^{2} (x) - 1

改写为

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

2 cos^{2} (x) - 1

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

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

将

1

改写为

1^{2}

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

使用指数法则:

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

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

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

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

使用平方差公式:

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

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

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

= sin (x) (2 cos (x) + 1) (2 cos (x) - 1)

sin (x) (2 cos (x) + 1) (2 cos (x) - 1) = 0

分别求解每个部分

sin (x) = 0 or 2 cos (x) + 1 = 0 or 2 cos (x) - 1 = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 0

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

2 cos (x) + 1 = 0 : x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

2 cos (x) + 1 = 0

将

1

到右边

2 cos (x) + 1 = 0

两边减去

1

2 cos (x) + 1 - 1 = 0 - 1

化简

2 cos (x) = - 1

2 cos (x) = - 1

两边除以

2

2 cos (x) = - 1

两边除以

2

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

化简

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

化简

\frac{2 cos ( x )}{2} : cos (x)

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

约分:

2

= cos (x)

化简

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

\frac{- 1}{2}

使用分式法则:

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

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

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

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

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

cos (x) = - \frac{2}{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{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

2 cos (x) - 1 = 0 : x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

2 cos (x) - 1 = 0

将

1

到右边

2 cos (x) - 1 = 0

两边加上

1

2 cos (x) - 1 + 1 = 0 + 1

化简

2 cos (x) = 1

2 cos (x) = 1

两边除以

2

2 cos (x) = 1

两边除以

2

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

化简

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

化简

\frac{2 cos ( x )}{2} : cos (x)

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

约分:

2

= cos (x)

化简

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

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

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

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

cos (x) = \frac{2}{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}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

合并所有解

x = 2 πn, x = π + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

流行的例子

cos (x) = 0.35

sin (\frac{17 π}{6})

cos (2 θ) = \frac{1}{2}

tan(arcsin(-4/5))

tan (arcsin (- \frac{4}{5}))

sin^{2} (\frac{π}{3})