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受欢迎的三角函数 >

cos^2(x)+3|cos(x)|-1=0

解答

cos^{2} (x) + 3 ∣ cos (x) ∣ - 1 = 0

解答

x = 1.87839 \dots + 2 πn, x = - 1.87839 \dots + 2 πn, x = 1.26319 \dots + 2 πn, x = 2 π - 1.26319 \dots + 2 πn

+1

度数

x = 107.62439 \dots^{\circ} + 36 0^{\circ} n, x = - 107.62439 \dots^{\circ} + 36 0^{\circ} n, x = 72.37560 \dots^{\circ} + 36 0^{\circ} n, x = 287.62439 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cos^{2} (x) + 3 ∣ cos (x) ∣ - 1 = 0

用替代法求解

cos^{2} (x) + 3 ∣ cos (x) ∣ - 1 = 0

令

: cos (x) = u

u^{2} + 3 ∣ u ∣ - 1 = 0

u^{2} + 3 ∣ u ∣ - 1 = 0 : u = \frac{3 - 13}{2} or u = \frac{- 3 + 13}{2}

u^{2} + 3 ∣ u ∣ - 1 = 0

找到正负区间

找到

∣ u ∣

的区间

u \geq 0

:

u \geq 0, ∣ u ∣ = u

对于

u \geq 0

改写

∣ u ∣ : ∣ u ∣ = u

使用绝对值运算法则: 若

u \geq 0

，则

∣ u ∣ = u

∣ u ∣ = u

u < 0

:

u < 0, ∣ u ∣ = - u

对于

u < 0

改写

∣ u ∣ : ∣ u ∣ = - u

使用绝对值运算法则: 若

u < 0

，则

∣ u ∣ = - u

∣ u ∣ = - u

确定区间:

u < 0, u \geq 0

∣ u ∣ u < 0 - u \geq 0 +

u < 0, u \geq 0

u < 0, u \geq 0

对每个区间解不等式

u < 0, u \geq 0

对于

u < 0 : u = \frac{3 - 13}{2}

对于

u < 0

改写

u^{2} + 3 ∣ u ∣ - 1 = 0

为

u^{2} + 3 (- u) - 1 = 0

u^{2} + 3 (- u) - 1 = 0 : u = \frac{3 + 13}{2}, u = \frac{3 - 13}{2}

u^{2} + 3 (- u) - 1 = 0

整理后得

u^{2} - 3 u - 1 = 0

使用求根公式求解

u^{2} - 3 u - 1 = 0

二次方程求根公式:

若

a = 1, b = - 3, c = - 1

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

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

(- 3)^{2} - 4 \cdot 1 \cdot (- 1) = 13

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 3^{2} + 4

3^{2} = 9

= 9 + 4

数字相加:

9 + 4 = 13

= 13

u_{1, 2} = \frac{- ( - 3 ) \pm 13}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 13}{2 \cdot 1}, u_{2} = \frac{- ( - 3 ) - 13}{2 \cdot 1}

u = \frac{- ( - 3 ) + 13}{2 \cdot 1} : \frac{3 + 13}{2}

\frac{- ( - 3 ) + 13}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{3 + 13}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{3 + 13}{2}

u = \frac{- ( - 3 ) - 13}{2 \cdot 1} : \frac{3 - 13}{2}

\frac{- ( - 3 ) - 13}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{3 - 13}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{3 - 13}{2}

二次方程组的解是:

u = \frac{3 + 13}{2}, u = \frac{3 - 13}{2}

合并区间

(u = \frac{3 - 13}{2} or u = \frac{3 + 13}{2}) and (u < 0)

合并重叠的区间

u = \frac{3 - 13}{2} or u = \frac{3 + 13}{2} and u < 0

两个区间的交集是指同时存在于这两个区间的数的集合

u = \frac{3 - 13}{2} or u = \frac{3 + 13}{2}

and

u < 0

u = \frac{3 - 13}{2}

u = \frac{3 - 13}{2}

对于

u \geq 0 : u = \frac{- 3 + 13}{2}

对于

u \geq 0

改写

u^{2} + 3 ∣ u ∣ - 1 = 0

为

u^{2} + 3 u - 1 = 0

u^{2} + 3 u - 1 = 0 : u = \frac{- 3 + 13}{2}, u = \frac{- 3 - 13}{2}

u^{2} + 3 u - 1 = 0

使用求根公式求解

u^{2} + 3 u - 1 = 0

二次方程求根公式:

若

a = 1, b = 3, c = - 1

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

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

3^{2} - 4 \cdot 1 \cdot (- 1) = 13

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

使用法则

- (- a) = a

= 3^{2} + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 3^{2} + 4

3^{2} = 9

= 9 + 4

数字相加:

9 + 4 = 13

= 13

u_{1, 2} = \frac{- 3 \pm 13}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 3 + 13}{2 \cdot 1}, u_{2} = \frac{- 3 - 13}{2 \cdot 1}

u = \frac{- 3 + 13}{2 \cdot 1} : \frac{- 3 + 13}{2}

\frac{- 3 + 13}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 3 + 13}{2}

u = \frac{- 3 - 13}{2 \cdot 1} : \frac{- 3 - 13}{2}

\frac{- 3 - 13}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 3 - 13}{2}

二次方程组的解是:

u = \frac{- 3 + 13}{2}, u = \frac{- 3 - 13}{2}

合并区间

(u = \frac{- 3 - 13}{2} or u = \frac{- 3 + 13}{2}) and (u \geq 0)

合并重叠的区间

u = \frac{- 3 - 13}{2} or u = \frac{- 3 + 13}{2} and u \geq 0

两个区间的交集是指同时存在于这两个区间的数的集合

u = \frac{- 3 - 13}{2} or u = \frac{- 3 + 13}{2}

and

u \geq 0

u = \frac{- 3 + 13}{2}

u = \frac{- 3 + 13}{2}

合并解:

u = \frac{3 - 13}{2} or u = \frac{- 3 + 13}{2}

u = \frac{3 - 13}{2} or u = \frac{- 3 + 13}{2}

u = cos (x)

代回

cos (x) = \frac{3 - 13}{2} or cos (x) = \frac{- 3 + 13}{2}

cos (x) = \frac{3 - 13}{2} or cos (x) = \frac{- 3 + 13}{2}

cos (x) = \frac{3 - 13}{2} : x = arccos (\frac{3 - 13}{2}) + 2 πn, x = - arccos (\frac{3 - 13}{2}) + 2 πn

cos (x) = \frac{3 - 13}{2}

使用反三角函数性质

cos (x) = \frac{3 - 13}{2}

cos (x) = \frac{3 - 13}{2}

的通解

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

x = arccos (\frac{3 - 13}{2}) + 2 πn, x = - arccos (\frac{3 - 13}{2}) + 2 πn

x = arccos (\frac{3 - 13}{2}) + 2 πn, x = - arccos (\frac{3 - 13}{2}) + 2 πn

cos (x) = \frac{- 3 + 13}{2} : x = arccos (\frac{- 3 + 13}{2}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 13}{2}) + 2 πn

cos (x) = \frac{- 3 + 13}{2}

使用反三角函数性质

cos (x) = \frac{- 3 + 13}{2}

cos (x) = \frac{- 3 + 13}{2}

的通解

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

x = arccos (\frac{- 3 + 13}{2}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 13}{2}) + 2 πn

x = arccos (\frac{- 3 + 13}{2}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 13}{2}) + 2 πn

合并所有解

x = arccos (\frac{3 - 13}{2}) + 2 πn, x = - arccos (\frac{3 - 13}{2}) + 2 πn, x = arccos (\frac{- 3 + 13}{2}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 13}{2}) + 2 πn

以小数形式表示解

x = 1.87839 \dots + 2 πn, x = - 1.87839 \dots + 2 πn, x = 1.26319 \dots + 2 πn, x = 2 π - 1.26319 \dots + 2 πn

作图

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