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受欢迎的三角函数 >

5cos^2(2x)+4cos^2(x)-5=0

解答

5 cos^{2} (2 x) + 4 cos^{2} (x) - 5 = 0

解答

x = 0.46364 \dots + 2 πn, x = 2 π - 0.46364 \dots + 2 πn, x = 2.67794 \dots + 2 πn, x = - 2.67794 \dots + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

+1

度数

x = 26.56505 \dots^{\circ} + 36 0^{\circ} n, x = 333.43494 \dots^{\circ} + 36 0^{\circ} n, x = 153.43494 \dots^{\circ} + 36 0^{\circ} n, x = - 153.43494 \dots^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

5 cos^{2} (2 x) + 4 cos^{2} (x) - 5 = 0

使用三角恒等式改写

- 5 + 4 cos^{2} (x) + 5 cos^{2} (2 x)

使用倍角公式:

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

= - 5 + 4 cos^{2} (x) + 5 (2 cos^{2} (x) - 1)^{2}

化简

- 5 + 4 cos^{2} (x) + 5 (2 cos^{2} (x) - 1)^{2} : 20 cos^{4} (x) - 16 cos^{2} (x)

- 5 + 4 cos^{2} (x) + 5 (2 cos^{2} (x) - 1)^{2}

(2 cos^{2} (x) - 1)^{2} : 4 cos^{4} (x) - 4 cos^{2} (x) + 1

使用完全平方公式:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

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

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

化简

(2 cos^{2} (x))^{2} - 2 \cdot 2 cos^{2} (x) \cdot 1 + 1^{2} : 4 cos^{4} (x) - 4 cos^{2} (x) + 1

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

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

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

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

(cos^{2} (x))^{2} : cos^{4} (x)

使用指数法则:

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

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

数字相乘:

2 \cdot 2 = 4

= cos^{4} (x)

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

2^{2} = 4

= 4 cos^{4} (x)

2 \cdot 2 cos^{2} (x) \cdot 1 = 4 cos^{2} (x)

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

数字相乘:

2 \cdot 2 \cdot 1 = 4

= 4 cos^{2} (x)

= 4 cos^{4} (x) - 4 cos^{2} (x) + 1

= 4 cos^{4} (x) - 4 cos^{2} (x) + 1

= - 5 + 4 cos^{2} (x) + 5 (4 cos^{4} (x) - 4 cos^{2} (x) + 1)

乘开

5 (4 cos^{4} (x) - 4 cos^{2} (x) + 1) : 20 cos^{4} (x) - 20 cos^{2} (x) + 5

5 (4 cos^{4} (x) - 4 cos^{2} (x) + 1)

打开括号

= 5 \cdot 4 cos^{4} (x) + 5 (- 4 cos^{2} (x)) + 5 \cdot 1

使用加减运算法则

+ (- a) = - a

= 5 \cdot 4 cos^{4} (x) - 5 \cdot 4 cos^{2} (x) + 5 \cdot 1

化简

5 \cdot 4 cos^{4} (x) - 5 \cdot 4 cos^{2} (x) + 5 \cdot 1 : 20 cos^{4} (x) - 20 cos^{2} (x) + 5

5 \cdot 4 cos^{4} (x) - 5 \cdot 4 cos^{2} (x) + 5 \cdot 1

数字相乘:

5 \cdot 4 = 20

= 20 cos^{4} (x) - 20 cos^{2} (x) + 5 \cdot 1

数字相乘:

5 \cdot 1 = 5

= 20 cos^{4} (x) - 20 cos^{2} (x) + 5

= 20 cos^{4} (x) - 20 cos^{2} (x) + 5

= - 5 + 4 cos^{2} (x) + 20 cos^{4} (x) - 20 cos^{2} (x) + 5

化简

- 5 + 4 cos^{2} (x) + 20 cos^{4} (x) - 20 cos^{2} (x) + 5 : 20 cos^{4} (x) - 16 cos^{2} (x)

- 5 + 4 cos^{2} (x) + 20 cos^{4} (x) - 20 cos^{2} (x) + 5

对同类项分组

= 4 cos^{2} (x) + 20 cos^{4} (x) - 20 cos^{2} (x) - 5 + 5

同类项相加:

4 cos^{2} (x) - 20 cos^{2} (x) = - 16 cos^{2} (x)

= - 16 cos^{2} (x) + 20 cos^{4} (x) - 5 + 5

- 5 + 5 = 0

= 20 cos^{4} (x) - 16 cos^{2} (x)

= 20 cos^{4} (x) - 16 cos^{2} (x)

= 20 cos^{4} (x) - 16 cos^{2} (x)

- 16 cos^{2} (x) + 20 cos^{4} (x) = 0

用替代法求解

- 16 cos^{2} (x) + 20 cos^{4} (x) = 0

令

: cos (x) = u

- 16 u^{2} + 20 u^{4} = 0

- 16 u^{2} + 20 u^{4} = 0 : u = \frac{2 5}{5}, u = - \frac{2 5}{5}, u = 0

- 16 u^{2} + 20 u^{4} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

20 u^{4} - 16 u^{2} = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

20 v^{2} - 16 v = 0

解

20 v^{2} - 16 v = 0 : v = \frac{4}{5}, v = 0

20 v^{2} - 16 v = 0

使用求根公式求解

20 v^{2} - 16 v = 0

二次方程求根公式:

若

a = 20, b = - 16, c = 0

v_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 20 \cdot 0}{2 \cdot 20}

v_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 20 \cdot 0}{2 \cdot 20}

(- 16)^{2} - 4 \cdot 20 \cdot 0 = 16

(- 16)^{2} - 4 \cdot 20 \cdot 0

使用指数法则:

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

若

n

是偶数

(- 16)^{2} = 1 6^{2}

= 1 6^{2} - 4 \cdot 20 \cdot 0

使用法则

0 \cdot a = 0

= 1 6^{2} - 0

1 6^{2} - 0 = 1 6^{2}

= 1 6^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 16

v_{1, 2} = \frac{- ( - 16 ) \pm 16}{2 \cdot 20}

将解分隔开

v_{1} = \frac{- ( - 16 ) + 16}{2 \cdot 20}, v_{2} = \frac{- ( - 16 ) - 16}{2 \cdot 20}

v = \frac{- ( - 16 ) + 16}{2 \cdot 20} : \frac{4}{5}

\frac{- ( - 16 ) + 16}{2 \cdot 20}

使用法则

- (- a) = a

= \frac{16 + 16}{2 \cdot 20}

数字相加:

16 + 16 = 32

= \frac{32}{2 \cdot 20}

数字相乘:

2 \cdot 20 = 40

= \frac{32}{40}

约分:

8

= \frac{4}{5}

v = \frac{- ( - 16 ) - 16}{2 \cdot 20} : 0

\frac{- ( - 16 ) - 16}{2 \cdot 20}

使用法则

- (- a) = a

= \frac{16 - 16}{2 \cdot 20}

数字相减:

16 - 16 = 0

= \frac{0}{2 \cdot 20}

数字相乘:

2 \cdot 20 = 40

= \frac{0}{40}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

二次方程组的解是:

v = \frac{4}{5}, v = 0

v = \frac{4}{5}, v = 0

代回

v = u^{2}

，求解

u

解

u^{2} = \frac{4}{5} : u = \frac{2 5}{5}, u = - \frac{2 5}{5}

u^{2} = \frac{4}{5}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = \frac{4}{5}, u = - \frac{4}{5}

\frac{4}{5} = \frac{2 5}{5}

\frac{4}{5}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{4}{5}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2}{5}

\frac{2}{5}

有理化:

\frac{2 5}{5}

\frac{2}{5}

乘以共轭根式

\frac{5}{5}

= \frac{2 5}{5 5}

55 = 5

55

使用根式运算法则:

a a = a

55 = 5

= 5

= \frac{2 5}{5}

= \frac{2 5}{5}

- \frac{4}{5} = - \frac{2 5}{5}

- \frac{4}{5}

化简

\frac{4}{5} : \frac{2}{5}

\frac{4}{5}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{4}{5}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2}{5}

= - \frac{2}{5}

- \frac{2}{5}

有理化:

- \frac{2 5}{5}

- \frac{2}{5}

乘以共轭根式

\frac{5}{5}

= - \frac{2 5}{5 5}

55 = 5

55

使用根式运算法则:

a a = a

55 = 5

= 5

= - \frac{2 5}{5}

= - \frac{2 5}{5}

u = \frac{2 5}{5}, u = - \frac{2 5}{5}

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

u = 0

解为

u = \frac{2 5}{5}, u = - \frac{2 5}{5}, u = 0

u = cos (x)

代回

cos (x) = \frac{2 5}{5}, cos (x) = - \frac{2 5}{5}, cos (x) = 0

cos (x) = \frac{2 5}{5}, cos (x) = - \frac{2 5}{5}, cos (x) = 0

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

cos (x) = \frac{2 5}{5}

使用反三角函数性质

cos (x) = \frac{2 5}{5}

cos (x) = \frac{2 5}{5}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{2 5}{5}) + 2 πn, x = 2 π - arccos (\frac{2 5}{5}) + 2 πn

x = arccos (\frac{2 5}{5}) + 2 πn, x = 2 π - arccos (\frac{2 5}{5}) + 2 πn

cos (x) = - \frac{2 5}{5} : x = arccos (- \frac{2 5}{5}) + 2 πn, x = - arccos (- \frac{2 5}{5}) + 2 πn

cos (x) = - \frac{2 5}{5}

使用反三角函数性质

cos (x) = - \frac{2 5}{5}

cos (x) = - \frac{2 5}{5}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{2 5}{5}) + 2 πn, x = - arccos (- \frac{2 5}{5}) + 2 πn

x = arccos (- \frac{2 5}{5}) + 2 πn, x = - arccos (- \frac{2 5}{5}) + 2 πn

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

cos (x) = 0

cos (x) = 0

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}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

合并所有解

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

以小数形式表示解

x = 0.46364 \dots + 2 πn, x = 2 π - 0.46364 \dots + 2 πn, x = 2.67794 \dots + 2 πn, x = - 2.67794 \dots + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

作图

流行的例子

sin(3x+10)=cos(x+24)

sin (3 x + 10) = cos (x + 2 4^{\circ})

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

((1+cos^2(a)))/(sin^2(a))= 5/3

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

d^2+13d+36=(sin^2(x))/2

d^{2} + 13 d + 36 = \frac{sin ^{2} ( x )}{2}

(tan^2(x)-4)/(cos(x)+5)=0

\frac{tan ^{2} ( x ) - 4}{cos ( x ) + 5} = 0