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受欢迎的三角函数 >

sqrt(2)cos(x)-sqrt(2)sin(x)=1,0<= x<2pi

解答

2 cos (x) - 2 sin (x) = 1, 0 \leq x < 2 π

解答

x = π + 1.30899 \dots, x = 0.26179 \dots

+1

度数

x = 25 5^{\circ}, x = 1 5^{\circ}

求解步骤

2 cos (x) - 2 sin (x) = 1, 0 \leq x < 2 π

两边加上

2 sin (x)

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

两边进行平方

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

两边减去

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

2 cos^{2} (x) - 1 - 22 sin (x) - 2 sin^{2} (x) = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

- 1 + 2 (1 - sin^{2} (x)) - 2 sin^{2} (x) - 2 sin (x) 2 : - 4 sin^{2} (x) - 22 sin (x) + 1

- 1 + 2 (1 - sin^{2} (x)) - 2 sin^{2} (x) - 2 sin (x) 2

= - 1 + 2 (1 - sin^{2} (x)) - 2 sin^{2} (x) - 22 sin (x)

乘开

2 (1 - sin^{2} (x)) : 2 - 2 sin^{2} (x)

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

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (x)

= 2 \cdot 1 - 2 sin^{2} (x)

数字相乘:

2 \cdot 1 = 2

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

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

化简

- 1 + 2 - 2 sin^{2} (x) - 2 sin^{2} (x) - 2 sin (x) 2 : - 4 sin^{2} (x) - 22 sin (x) + 1

- 1 + 2 - 2 sin^{2} (x) - 2 sin^{2} (x) - 2 sin (x) 2

同类项相加:

- 2 sin^{2} (x) - 2 sin^{2} (x) = - 4 sin^{2} (x)

= - 1 + 2 - 4 sin^{2} (x) - 22 sin (x)

数字相加/相减:

- 1 + 2 = 1

= - 4 sin^{2} (x) - 22 sin (x) + 1

= - 4 sin^{2} (x) - 22 sin (x) + 1

= - 4 sin^{2} (x) - 22 sin (x) + 1

1 - 4 sin^{2} (x) - 2 sin (x) 2 = 0

用替代法求解

1 - 4 sin^{2} (x) - 2 sin (x) 2 = 0

令

: sin (x) = u

1 - 4 u^{2} - 2 u 2 = 0

1 - 4 u^{2} - 2 u 2 = 0 : u = - \frac{2 + 6}{4}, u = \frac{6 - 2}{4}

1 - 4 u^{2} - 2 u 2 = 0

改写成标准形式

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

- 4 u^{2} - 22 u + 1 = 0

使用求根公式求解

- 4 u^{2} - 22 u + 1 = 0

二次方程求根公式:

若

a = - 4, b = - 22, c = 1

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

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

(- 22)^{2} - 4 (- 4) \cdot 1 = 26

(- 22)^{2} - 4 (- 4) \cdot 1

使用法则

- (- a) = a

= (- 22)^{2} + 4 \cdot 4 \cdot 1

(- 22)^{2} = 2^{3}

(- 22)^{2}

使用指数法则:

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

若

n

是偶数

(- 22)^{2} = (22)^{2}

= (22)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} (2)^{2}

(2)^{2} : 2

使用根式运算法则:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

约分:

2

= 1

= 2

= 2^{2} \cdot 2

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

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

= 2^{2 + 1}

数字相加:

2 + 1 = 3

= 2^{3}

4 \cdot 4 \cdot 1 = 16

4 \cdot 4 \cdot 1

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 16

= 2^{3} + 16

2^{3} = 8

= 8 + 16

数字相加:

8 + 16 = 24

= 24

24

质因数分解:

2^{3} \cdot 3

24

24

除以

224 = 12 \cdot 2

= 2 \cdot 12

12

除以

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 3

= 2^{3} \cdot 3

= 2^{3} \cdot 3

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2 \cdot 3

使用根式运算法则:

n ab = n a n b

= 2^{2} 2 \cdot 3

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 2 \cdot 3

整理后得

= 26

u_{1, 2} = \frac{- ( - 2 2 ) \pm 2 6}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- ( - 2 2 ) + 2 6}{2 ( - 4 )}, u_{2} = \frac{- ( - 2 2 ) - 2 6}{2 ( - 4 )}

u = \frac{- ( - 2 2 ) + 2 6}{2 ( - 4 )} : - \frac{2 + 6}{4}

\frac{- ( - 2 2 ) + 2 6}{2 ( - 4 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{2 2 + 2 6}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{2 2 + 2 6}{- 8}

使用分式法则:

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

= - \frac{2 2 + 2 6}{8}

消掉

\frac{2 2 + 2 6}{8} : \frac{2 + 6}{4}

\frac{2 2 + 2 6}{8}

因式分解出通项

2

= \frac{2 ( 2 + 6 )}{8}

约分:

2

= \frac{2 + 6}{4}

= - \frac{2 + 6}{4}

u = \frac{- ( - 2 2 ) - 2 6}{2 ( - 4 )} : \frac{6 - 2}{4}

\frac{- ( - 2 2 ) - 2 6}{2 ( - 4 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{2 2 - 2 6}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

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

使用分式法则:

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

22 - 26 = - (26 - 22)

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

因式分解出通项

2

= \frac{2 ( 6 - 2 )}{8}

约分:

2

= \frac{6 - 2}{4}

二次方程组的解是:

u = - \frac{2 + 6}{4}, u = \frac{6 - 2}{4}

u = sin (x)

代回

sin (x) = - \frac{2 + 6}{4}, sin (x) = \frac{6 - 2}{4}

sin (x) = - \frac{2 + 6}{4}, sin (x) = \frac{6 - 2}{4}

sin (x) = - \frac{2 + 6}{4}, 0 \leq x < 2 π : x = π + arcsin (\frac{2 + 6}{4}), x = - arcsin (\frac{2 + 6}{4}) + 2 π

sin (x) = - \frac{2 + 6}{4}, 0 \leq x < 2 π

使用反三角函数性质

sin (x) = - \frac{2 + 6}{4}

sin (x) = - \frac{2 + 6}{4}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{2 + 6}{4}) + 2 πn, x = π + arcsin (\frac{2 + 6}{4}) + 2 πn

x = arcsin (- \frac{2 + 6}{4}) + 2 πn, x = π + arcsin (\frac{2 + 6}{4}) + 2 πn

在

0 \leq x < 2 π

范围内的解

x = π + arcsin (\frac{2 + 6}{4}), x = - arcsin (\frac{2 + 6}{4}) + 2 π

sin (x) = \frac{6 - 2}{4}, 0 \leq x < 2 π : x = arcsin (\frac{6 - 2}{4}), x = π - arcsin (\frac{6 - 2}{4})

sin (x) = \frac{6 - 2}{4}, 0 \leq x < 2 π

使用反三角函数性质

sin (x) = \frac{6 - 2}{4}

sin (x) = \frac{6 - 2}{4}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{6 - 2}{4}) + 2 πn, x = π - arcsin (\frac{6 - 2}{4}) + 2 πn

x = arcsin (\frac{6 - 2}{4}) + 2 πn, x = π - arcsin (\frac{6 - 2}{4}) + 2 πn

在

0 \leq x < 2 π

范围内的解

x = arcsin (\frac{6 - 2}{4}), x = π - arcsin (\frac{6 - 2}{4})

合并所有解

x = π + arcsin (\frac{2 + 6}{4}), x = - arcsin (\frac{2 + 6}{4}) + 2 π, x = arcsin (\frac{6 - 2}{4}), x = π - arcsin (\frac{6 - 2}{4})

将解代入原方程进行验证

将它们代入

2 cos (x) - 2 sin (x) = 1

检验解是否符合

去除与方程不符的解。

检验

π + arcsin (\frac{2 + 6}{4})

的解:

真

π + arcsin (\frac{2 + 6}{4})

代入

n = 1

π + arcsin (\frac{2 + 6}{4})

对于

2 cos (x) - 2 sin (x) = 1

代入

x = π + arcsin (\frac{2 + 6}{4})

2 cos (π + arcsin (\frac{2 + 6}{4})) - 2 sin (π + arcsin (\frac{2 + 6}{4})) = 1

整理后得

1 = 1

\Rightarrow 真

检验

- arcsin (\frac{2 + 6}{4}) + 2 π

的解:

假

- arcsin (\frac{2 + 6}{4}) + 2 π

代入

n = 1

- arcsin (\frac{2 + 6}{4}) + 2 π

对于

2 cos (x) - 2 sin (x) = 1

代入

x = - arcsin (\frac{2 + 6}{4}) + 2 π

2 cos (- arcsin (\frac{2 + 6}{4}) + 2 π) - 2 sin (- arcsin (\frac{2 + 6}{4}) + 2 π) = 1

整理后得

1.73205 \dots = 1

\Rightarrow 假

检验

arcsin (\frac{6 - 2}{4})

的解:

真

arcsin (\frac{6 - 2}{4})

代入

n = 1

arcsin (\frac{6 - 2}{4})

对于

2 cos (x) - 2 sin (x) = 1

代入

x = arcsin (\frac{6 - 2}{4})

2 cos (arcsin (\frac{6 - 2}{4})) - 2 sin (arcsin (\frac{6 - 2}{4})) = 1

整理后得

1 = 1

\Rightarrow 真

检验

π - arcsin (\frac{6 - 2}{4})

的解:

假

π - arcsin (\frac{6 - 2}{4})

代入

n = 1

π - arcsin (\frac{6 - 2}{4})

对于

2 cos (x) - 2 sin (x) = 1

代入

x = π - arcsin (\frac{6 - 2}{4})

2 cos (π - arcsin (\frac{6 - 2}{4})) - 2 sin (π - arcsin (\frac{6 - 2}{4})) = 1

整理后得

- 1.73205 \dots = 1

\Rightarrow 假

x = π + arcsin (\frac{2 + 6}{4}), x = arcsin (\frac{6 - 2}{4})

以小数形式表示解

x = π + 1.30899 \dots, x = 0.26179 \dots

作图

流行的例子

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2 sin^{2} (x) - 3 cos (x) = 3

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sin^2(θ)-3sin(θ)=0

sin^{2} (θ) - 3 sin (θ) = 0

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