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

解答

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

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

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

两边减去

2 cos (x)

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

用 sin, cos 表示

- 1 - 2 cos (x) + cos (x) csc^{2} (x) tan^{2} (x)

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

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

使用基本三角恒等式:

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

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

化简

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

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

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

cos (x) (\frac{1}{sin ( x )})^{2} (\frac{sin ( x )}{cos ( x )})^{2}

(\frac{1}{sin ( x )})^{2} = \frac{1}{sin ^{2} ( x )}

(\frac{1}{sin ( x )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{sin ^{2} ( x )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( x )}

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

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

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

使用指数法则:

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

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

= \frac{1}{sin ^{2} ( x )} \cdot \frac{sin ^{2} ( x )}{cos ^{2} ( x )} cos (x)

分式相乘:

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

= \frac{1 \cdot sin ^{2} ( x ) cos ( x )}{sin ^{2} ( x ) cos ^{2} ( x )}

约分:

sin^{2} (x)

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

乘以:

1 \cdot cos (x) = cos (x)

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

约分:

cos (x)

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

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

将项转换为分式:

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

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

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

- 1 \cdot cos (x) - 2 cos (x) cos (x) + 1

1 \cdot cos (x) = cos (x)

1 \cdot cos (x)

乘以:

1 \cdot cos (x) = cos (x)

= cos (x)

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

2 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)

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

数字相加:

1 + 1 = 2

= 2 cos^{2} (x)

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

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

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

\frac{1 - cos ( x ) - 2 cos ^{2} ( x )}{cos ( x )} = 0

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

1 - cos (x) - 2 cos^{2} (x) = 0

用替代法求解

1 - cos (x) - 2 cos^{2} (x) = 0

令

: cos (x) = u

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

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

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

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

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

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 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:

3^{2} = 3

= 3

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

将解分隔开

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

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

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

去除括号:

(- a) = - a, - (- a) = a

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

数字相加:

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

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

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

去除括号:

(- a) = - a, - (- a) = a

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

数字相减:

1 - 3 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

使用分式法则:

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

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

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

u = cos (x)

代回

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

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

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

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

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

合并所有解

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

因为方程对以下值无定义:

π + 2 πn

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

流行的例子

sin (a) = \frac{4}{6}

cos (a) = 0.94

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

2 cos (2 x) - 1 = 0

2cos^2(t)-cos(t)-3=0

2 cos^{2} (t) - cos (t) - 3 = 0

2cos(4θ+20)=0.8

2 cos (4 θ + 20) = 0.8