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受欢迎的三角函数 >

sqrt(cos(θ))=2cos(θ)-1

解答

cos (θ) = 2 cos (θ) - 1

解答

θ = 2 πn

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n

求解步骤

cos (θ) = 2 cos (θ) - 1

用替代法求解

cos (θ) = 2 cos (θ) - 1

令

: cos (θ) = u

u = 2 u - 1

u = 2 u - 1 : u = 1

u = 2 u - 1

两边进行平方:

u = 4 u^{2} - 4 u + 1

u = 2 u - 1

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

展开

(u)^{2} : u

(u)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (u^{\frac{1}{2}})^{2}

使用指数法则:

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

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

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= u

展开

(2 u - 1)^{2} : 4 u^{2} - 4 u + 1

(2 u - 1)^{2}

使用完全平方公式:

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

a = 2 u, b = 1

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

化简

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

(2 u)^{2} - 2 \cdot 2 u \cdot 1 + 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

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

(2 u)^{2} = 4 u^{2}

(2 u)^{2}

使用指数法则:

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

= 2^{2} u^{2}

2^{2} = 4

= 4 u^{2}

2 \cdot 2 u \cdot 1 = 4 u

2 \cdot 2 u \cdot 1

数字相乘:

2 \cdot 2 \cdot 1 = 4

= 4 u

= 4 u^{2} - 4 u + 1

= 4 u^{2} - 4 u + 1

u = 4 u^{2} - 4 u + 1

u = 4 u^{2} - 4 u + 1

解

u = 4 u^{2} - 4 u + 1 : u = 1, u = \frac{1}{4}

u = 4 u^{2} - 4 u + 1

交换两边

4 u^{2} - 4 u + 1 = u

将

u

para o lado esquerdo

4 u^{2} - 4 u + 1 = u

两边减去

u

4 u^{2} - 4 u + 1 - u = u - u

化简

4 u^{2} - 5 u + 1 = 0

4 u^{2} - 5 u + 1 = 0

使用求根公式求解

4 u^{2} - 5 u + 1 = 0

二次方程求根公式:

若

a = 4, b = - 5, c = 1

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

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

(- 5)^{2} - 4 \cdot 4 \cdot 1 = 3

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

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 4 \cdot 1

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

数字相减:

25 - 16 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

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

将解分隔开

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

u = \frac{- ( - 5 ) + 3}{2 \cdot 4} : 1

\frac{- ( - 5 ) + 3}{2 \cdot 4}

使用法则

- (- a) = a

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

数字相加:

5 + 3 = 8

= \frac{8}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{8}{8}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 5 ) - 3}{2 \cdot 4} : \frac{1}{4}

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

使用法则

- (- a) = a

= \frac{5 - 3}{2 \cdot 4}

数字相减:

5 - 3 = 2

= \frac{2}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{2}{8}

约分:

2

= \frac{1}{4}

二次方程组的解是:

u = 1, u = \frac{1}{4}

u = 1, u = \frac{1}{4}

验证解:

u = 1

真

, u = \frac{1}{4}

假

将它们代入

u = 2 u - 1

检验解是否符合

去除与方程不符的解。

代入

u = 1 :

真

1 = 2 \cdot 1 - 1

1 = 1

1

使用法则

1 = 1

= 1

2 \cdot 1 - 1 = 1

2 \cdot 1 - 1

数字相乘:

2 \cdot 1 = 2

= 2 - 1

数字相减:

2 - 1 = 1

= 1

1 = 1

真

代入

u = \frac{1}{4} :

假

\frac{1}{4} = 2 (\frac{1}{4}) - 1

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

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{1}{4}

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{1}{2}

使用法则

1 = 1

= \frac{1}{2}

2 (\frac{1}{4}) - 1 = - \frac{1}{2}

2 (\frac{1}{4}) - 1

去除括号:

(a) = a

= 2 \cdot \frac{1}{4} - 1

2 \cdot \frac{1}{4} = \frac{1}{2}

2 \cdot \frac{1}{4}

分式相乘:

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

= \frac{1 \cdot 2}{4}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{4}

约分:

2

= \frac{1}{2}

= \frac{1}{2} - 1

将项转换为分式:

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

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

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 2 + 1}{2}

- 1 \cdot 2 + 1 = - 1

- 1 \cdot 2 + 1

数字相乘:

1 \cdot 2 = 2

= - 2 + 1

数字相加/相减:

- 2 + 1 = - 1

= - 1

= \frac{- 1}{2}

使用分式法则:

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

= - \frac{1}{2}

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

假

解是

u = 1

u = cos (θ)

代回

cos (θ) = 1

cos (θ) = 1

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

cos (θ) = 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}

θ = 0 + 2 πn

θ = 0 + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

合并所有解

θ = 2 πn

作图

流行的例子

2cos^2(2β)+cos(2β)-1=0,-360<= β<= 360

2 cos^{2} (2 β) + cos (2 β) - 1 = 0, - 36 0^{\circ} \leq β \leq 36 0^{\circ}

cos (2 x) = \frac{3}{7}

0 = 1 - 3 cos (θ)

(tan((3a)/2))tan(a/2)=3

(tan (\frac{3 a}{2})) tan (\frac{a}{2}) = 3

1=sqrt(3)sin(x)

1 = 3 sin (x)