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受欢迎的三角函数 >

7cos^2(θ)-3=-2cos(θ)+2

解答

7 cos^{2} (θ) - 3 = - 2 cos (θ) + 2

解答

θ = 0.77519 \dots + 2 πn, θ = 2 π - 0.77519 \dots + 2 πn, θ = π + 2 πn

+1

度数

θ = 44.41530 \dots^{\circ} + 36 0^{\circ} n, θ = 315.58469 \dots^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

7 cos^{2} (θ) - 3 = - 2 cos (θ) + 2

用替代法求解

7 cos^{2} (θ) - 3 = - 2 cos (θ) + 2

令

: cos (θ) = u

7 u^{2} - 3 = - 2 u + 2

7 u^{2} - 3 = - 2 u + 2 : u = \frac{5}{7}, u = - 1

7 u^{2} - 3 = - 2 u + 2

将

2

para o lado esquerdo

7 u^{2} - 3 = - 2 u + 2

两边减去

2

7 u^{2} - 3 - 2 = - 2 u + 2 - 2

化简

7 u^{2} - 5 = - 2 u

7 u^{2} - 5 = - 2 u

将

2 u

para o lado esquerdo

7 u^{2} - 5 = - 2 u

两边加上

2 u

7 u^{2} - 5 + 2 u = - 2 u + 2 u

化简

7 u^{2} - 5 + 2 u = 0

7 u^{2} - 5 + 2 u = 0

改写成标准形式

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

7 u^{2} + 2 u - 5 = 0

使用求根公式求解

7 u^{2} + 2 u - 5 = 0

二次方程求根公式:

若

a = 7, b = 2, c = - 5

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 7 ( - 5 )}{2 \cdot 7}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 7 ( - 5 )}{2 \cdot 7}

2^{2} - 4 \cdot 7 (- 5) = 12

2^{2} - 4 \cdot 7 (- 5)

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 7 \cdot 5

数字相乘:

4 \cdot 7 \cdot 5 = 140

= 2^{2} + 140

2^{2} = 4

= 4 + 140

数字相加:

4 + 140 = 144

= 144

因式分解数字:

144 = 1 2^{2}

= 1 2^{2}

使用根式运算法则:

n a^{n} = a

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- 2 \pm 12}{2 \cdot 7}

将解分隔开

u_{1} = \frac{- 2 + 12}{2 \cdot 7}, u_{2} = \frac{- 2 - 12}{2 \cdot 7}

u = \frac{- 2 + 12}{2 \cdot 7} : \frac{5}{7}

\frac{- 2 + 12}{2 \cdot 7}

数字相加/相减:

- 2 + 12 = 10

= \frac{10}{2 \cdot 7}

数字相乘:

2 \cdot 7 = 14

= \frac{10}{14}

约分:

2

= \frac{5}{7}

u = \frac{- 2 - 12}{2 \cdot 7} : - 1

\frac{- 2 - 12}{2 \cdot 7}

数字相减:

- 2 - 12 = - 14

= \frac{- 14}{2 \cdot 7}

数字相乘:

2 \cdot 7 = 14

= \frac{- 14}{14}

使用分式法则:

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

= - \frac{14}{14}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = \frac{5}{7}, u = - 1

u = cos (θ)

代回

cos (θ) = \frac{5}{7}, cos (θ) = - 1

cos (θ) = \frac{5}{7}, cos (θ) = - 1

cos (θ) = \frac{5}{7} : θ = arccos (\frac{5}{7}) + 2 πn, θ = 2 π - arccos (\frac{5}{7}) + 2 πn

cos (θ) = \frac{5}{7}

使用反三角函数性质

cos (θ) = \frac{5}{7}

cos (θ) = \frac{5}{7}

的通解

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

θ = arccos (\frac{5}{7}) + 2 πn, θ = 2 π - arccos (\frac{5}{7}) + 2 πn

θ = arccos (\frac{5}{7}) + 2 πn, θ = 2 π - arccos (\frac{5}{7}) + 2 πn

cos (θ) = - 1 : θ = π + 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 = π + 2 πn

θ = π + 2 πn

合并所有解

θ = arccos (\frac{5}{7}) + 2 πn, θ = 2 π - arccos (\frac{5}{7}) + 2 πn, θ = π + 2 πn

以小数形式表示解

θ = 0.77519 \dots + 2 πn, θ = 2 π - 0.77519 \dots + 2 πn, θ = π + 2 πn

作图

流行的例子

(sin(2x))/2 =sin(x)

\frac{sin ( 2 x )}{2} = sin (x)

sin^2(θ)+sin(θ)-6=0

sin^{2} (θ) + sin (θ) - 6 = 0

solvefor y,x+tan(xy)=0

so l v e f or y, x + tan (x y) = 0

- sin (2 θ) = 0

∣ cos (2 x) ∣ = 1