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受欢迎的三角函数 >

8cos^2(θ)-3cos(θ)=0,0<= θ<= 180

解答

8 cos^{2} (θ) - 3 cos (θ) = 0, 0^{\circ} \leq θ \leq 18 0^{\circ}

解答

θ = 1.18639 \dots, θ = 9 0^{\circ}

+1

弧度

θ = 1.18639 \dots, θ = \frac{π}{2}

求解步骤

8 cos^{2} (θ) - 3 cos (θ) = 0, 0^{\circ} \leq θ \leq 18 0^{\circ}

用替代法求解

8 cos^{2} (θ) - 3 cos (θ) = 0

令

: cos (θ) = u

8 u^{2} - 3 u = 0

8 u^{2} - 3 u = 0 : u = \frac{3}{8}, u = 0

8 u^{2} - 3 u = 0

使用求根公式求解

8 u^{2} - 3 u = 0

二次方程求根公式:

若

a = 8, b = - 3, c = 0

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 8 \cdot 0}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 8 \cdot 0}{2 \cdot 8}

(- 3)^{2} - 4 \cdot 8 \cdot 0 = 3

(- 3)^{2} - 4 \cdot 8 \cdot 0

使用指数法则:

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

若

n

是偶数

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

= 3^{2} - 4 \cdot 8 \cdot 0

使用法则

0 \cdot a = 0

= 3^{2} - 0

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

= 3^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 3

u_{1, 2} = \frac{- ( - 3 ) \pm 3}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 3}{2 \cdot 8}, u_{2} = \frac{- ( - 3 ) - 3}{2 \cdot 8}

u = \frac{- ( - 3 ) + 3}{2 \cdot 8} : \frac{3}{8}

\frac{- ( - 3 ) + 3}{2 \cdot 8}

使用法则

- (- a) = a

= \frac{3 + 3}{2 \cdot 8}

数字相加:

3 + 3 = 6

= \frac{6}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{6}{16}

约分:

2

= \frac{3}{8}

u = \frac{- ( - 3 ) - 3}{2 \cdot 8} : 0

\frac{- ( - 3 ) - 3}{2 \cdot 8}

使用法则

- (- a) = a

= \frac{3 - 3}{2 \cdot 8}

数字相减:

3 - 3 = 0

= \frac{0}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{0}{16}

使用法则

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

= 0

二次方程组的解是:

u = \frac{3}{8}, u = 0

u = cos (θ)

代回

cos (θ) = \frac{3}{8}, cos (θ) = 0

cos (θ) = \frac{3}{8}, cos (θ) = 0

cos (θ) = \frac{3}{8}, 0 \leq θ \leq 18 0^{\circ} : θ = arccos (\frac{3}{8})

cos (θ) = \frac{3}{8}, 0 \leq θ \leq 18 0^{\circ}

使用反三角函数性质

cos (θ) = \frac{3}{8}

cos (θ) = \frac{3}{8}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

θ = arccos (\frac{3}{8}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{3}{8}) + 36 0^{\circ} n

θ = arccos (\frac{3}{8}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{3}{8}) + 36 0^{\circ} n

在

0 \leq θ \leq 18 0^{\circ}

范围内的解

θ = arccos (\frac{3}{8})

cos (θ) = 0, 0 \leq θ \leq 18 0^{\circ} : θ = 9 0^{\circ}

cos (θ) = 0, 0 \leq θ \leq 18 0^{\circ}

cos (θ) = 0

的通解

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

在

0 \leq θ \leq 18 0^{\circ}

范围内的解

θ = 9 0^{\circ}

合并所有解

θ = arccos (\frac{3}{8}), θ = 9 0^{\circ}

以小数形式表示解

θ = 1.18639 \dots, θ = 9 0^{\circ}

作图

流行的例子

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tan (x + 2) = cot (4 x + 28)

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