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受欢迎的三角函数 >

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

解答

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

解答

x = 2.23703 \dots + 2 πn, x = - 2.23703 \dots + 2 πn

+1

度数

x = 128.17270 \dots^{\circ} + 36 0^{\circ} n, x = - 128.17270 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: cos (x) = u

u^{2} = u + 1

u^{2} = u + 1 : u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

u^{2} = u + 1

将

1

para o lado esquerdo

u^{2} = u + 1

两边减去

1

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

化简

u^{2} - 1 = u

u^{2} - 1 = u

将

u

para o lado esquerdo

u^{2} - 1 = u

两边减去

u

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

化简

u^{2} - 1 - u = 0

u^{2} - 1 - u = 0

改写成标准形式

a x^{2} + b x + c = 0

u^{2} - u - 1 = 0

使用求根公式求解

u^{2} - u - 1 = 0

二次方程求根公式:

若

a = 1, b = - 1, c = - 1

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

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

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

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

使用法则

- (- a) = a

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

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

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

将解分隔开

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

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

\frac{- ( - 1 ) + 5}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 + 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 5}{2}

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

\frac{- ( - 1 ) - 5}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 - 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 5}{2}

二次方程组的解是:

u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

u = cos (x)

代回

cos (x) = \frac{1 + 5}{2}, cos (x) = \frac{1 - 5}{2}

cos (x) = \frac{1 + 5}{2}, cos (x) = \frac{1 - 5}{2}

cos (x) = \frac{1 + 5}{2} :

无解

cos (x) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

无解

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

cos (x) = \frac{1 - 5}{2}

使用反三角函数性质

cos (x) = \frac{1 - 5}{2}

cos (x) = \frac{1 - 5}{2}

的通解

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

x = arccos (\frac{1 - 5}{2}) + 2 πn, x = - arccos (\frac{1 - 5}{2}) + 2 πn

x = arccos (\frac{1 - 5}{2}) + 2 πn, x = - arccos (\frac{1 - 5}{2}) + 2 πn

合并所有解

x = arccos (\frac{1 - 5}{2}) + 2 πn, x = - arccos (\frac{1 - 5}{2}) + 2 πn

以小数形式表示解

x = 2.23703 \dots + 2 πn, x = - 2.23703 \dots + 2 πn

作图

流行的例子

6sin^2(x)=7-5cos(x)

6 sin^{2} (x) = 7 - 5 cos (x)

3+3cos(x)=2(1-cos^2(x))

3 + 3 cos (x) = 2 (1 - cos^{2} (x))

sin (x) = 1.5

cot^2(θ)-4csc(θ)=-5

cot^{2} (θ) - 4 csc (θ) = - 5

tan (θ) = \frac{2}{7}