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受欢迎的三角函数 >

2+7cos(θ)=4sin^2(θ),-180<= x<= 180

解答

2 + 7 cos (θ) = 4 sin^{2} (θ), - 18 0^{\circ} \leq x \leq 18 0^{\circ}

解答

θ = 1.31811 \dots, θ = - 1.31811 \dots

求解步骤

2 + 7 cos (θ) = 4 sin^{2} (θ), - 18 0^{\circ} \leq x \leq 18 0^{\circ}

两边减去

4 sin^{2} (θ)

2 + 7 cos (θ) - 4 sin^{2} (θ) = 0

使用三角恒等式改写

2 - 4 sin^{2} (θ) + 7 cos (θ)

使用毕达哥拉斯恒等式:

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

sin^{2} (x) = 1 - cos^{2} (x)

= 2 - 4 (1 - cos^{2} (θ)) + 7 cos (θ)

化简

2 - 4 (1 - cos^{2} (θ)) + 7 cos (θ) : 4 cos^{2} (θ) + 7 cos (θ) - 2

2 - 4 (1 - cos^{2} (θ)) + 7 cos (θ)

乘开

- 4 (1 - cos^{2} (θ)) : - 4 + 4 cos^{2} (θ)

- 4 (1 - cos^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 4, b = 1, c = cos^{2} (θ)

= - 4 \cdot 1 - (- 4) cos^{2} (θ)

使用加减运算法则

- (- a) = a

= - 4 \cdot 1 + 4 cos^{2} (θ)

数字相乘:

4 \cdot 1 = 4

= - 4 + 4 cos^{2} (θ)

= 2 - 4 + 4 cos^{2} (θ) + 7 cos (θ)

数字相减:

2 - 4 = - 2

= 4 cos^{2} (θ) + 7 cos (θ) - 2

= 4 cos^{2} (θ) + 7 cos (θ) - 2

- 2 + 4 cos^{2} (θ) + 7 cos (θ) = 0

用替代法求解

- 2 + 4 cos^{2} (θ) + 7 cos (θ) = 0

令

: cos (θ) = u

- 2 + 4 u^{2} + 7 u = 0

- 2 + 4 u^{2} + 7 u = 0 : u = \frac{1}{4}, u = - 2

- 2 + 4 u^{2} + 7 u = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 4, b = 7, c = - 2

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

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

7^{2} - 4 \cdot 4 (- 2) = 9

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 4 \cdot 2 = 32

= 7^{2} + 32

7^{2} = 49

= 49 + 32

数字相加:

49 + 32 = 81

= 81

因式分解数字:

81 = 9^{2}

= 9^{2}

使用根式运算法则:

n a^{n} = a

9^{2} = 9

= 9

u_{1, 2} = \frac{- 7 \pm 9}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- 7 + 9}{2 \cdot 4}, u_{2} = \frac{- 7 - 9}{2 \cdot 4}

u = \frac{- 7 + 9}{2 \cdot 4} : \frac{1}{4}

\frac{- 7 + 9}{2 \cdot 4}

数字相加/相减:

- 7 + 9 = 2

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

数字相乘:

2 \cdot 4 = 8

= \frac{2}{8}

约分:

2

= \frac{1}{4}

u = \frac{- 7 - 9}{2 \cdot 4} : - 2

\frac{- 7 - 9}{2 \cdot 4}

数字相减:

- 7 - 9 = - 16

= \frac{- 16}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 16}{8}

使用分式法则:

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

= - \frac{16}{8}

数字相除:

\frac{16}{8} = 2

= - 2

二次方程组的解是:

u = \frac{1}{4}, u = - 2

u = cos (θ)

代回

cos (θ) = \frac{1}{4}, cos (θ) = - 2

cos (θ) = \frac{1}{4}, cos (θ) = - 2

cos (θ) = \frac{1}{4}, - 18 0^{\circ} \leq x \leq 18 0^{\circ} : θ = arccos (\frac{1}{4}), θ = - arccos (\frac{1}{4})

cos (θ) = \frac{1}{4}, - 18 0^{\circ} \leq x \leq 18 0^{\circ}

使用反三角函数性质

cos (θ) = \frac{1}{4}

cos (θ) = \frac{1}{4}

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{1}{4}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{1}{4}) + 36 0^{\circ} n

θ = arccos (\frac{1}{4}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{1}{4}) + 36 0^{\circ} n

在

- 18 0^{\circ} \leq x \leq 18 0^{\circ}

范围内的解

θ = arccos (\frac{1}{4}), θ = - arccos (\frac{1}{4})

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

无解

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

- 1 \leq cos (x) \leq 1

无解

合并所有解

θ = arccos (\frac{1}{4}), θ = - arccos (\frac{1}{4})

以小数形式表示解

θ = 1.31811 \dots, θ = - 1.31811 \dots

作图

流行的例子

1 = 2 cos (3 θ)

cos(4θ)+cos(2θ)=cos(θ)

cos (4 θ) + cos (2 θ) = cos (θ)

(tan(x)+3)(tan(x)+2)=0

(tan (x) + 3) (tan (x) + 2) = 0

2tan(θ)+sqrt(6)=0

2 tan (θ) + 6 = 0

8 = 8 cos (3 θ)