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受欢迎的三角函数 >

sqrt(6)sin(x)-sqrt(2)cos(x)=sqrt(2)

解答

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

解答

x = π + 2 πn, x = \frac{π}{3} + 2 πn

+1

度数

x = 18 0^{\circ} + 36 0^{\circ} n, x = 6 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边加上

2 cos (x)

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

两边进行平方

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

两边减去

(2 + 2 cos (x))^{2}

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

使用三角恒等式改写

- 2 - 2 cos^{2} (x) - 4 cos (x) + 6 sin^{2} (x)

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 - 2 cos^{2} (x) - 4 cos (x) + 6 (1 - cos^{2} (x))

化简

- 2 - 2 cos^{2} (x) - 4 cos (x) + 6 (1 - cos^{2} (x)) : - 8 cos^{2} (x) - 4 cos (x) + 4

- 2 - 2 cos^{2} (x) - 4 cos (x) + 6 (1 - cos^{2} (x))

乘开

6 (1 - cos^{2} (x)) : 6 - 6 cos^{2} (x)

6 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 6, b = 1, c = cos^{2} (x)

= 6 \cdot 1 - 6 cos^{2} (x)

数字相乘:

6 \cdot 1 = 6

= 6 - 6 cos^{2} (x)

= - 2 - 2 cos^{2} (x) - 4 cos (x) + 6 - 6 cos^{2} (x)

化简

- 2 - 2 cos^{2} (x) - 4 cos (x) + 6 - 6 cos^{2} (x) : - 8 cos^{2} (x) - 4 cos (x) + 4

- 2 - 2 cos^{2} (x) - 4 cos (x) + 6 - 6 cos^{2} (x)

对同类项分组

= - 2 cos^{2} (x) - 4 cos (x) - 6 cos^{2} (x) - 2 + 6

同类项相加:

- 2 cos^{2} (x) - 6 cos^{2} (x) = - 8 cos^{2} (x)

= - 8 cos^{2} (x) - 4 cos (x) - 2 + 6

数字相加/相减:

- 2 + 6 = 4

= - 8 cos^{2} (x) - 4 cos (x) + 4

= - 8 cos^{2} (x) - 4 cos (x) + 4

= - 8 cos^{2} (x) - 4 cos (x) + 4

4 - 4 cos (x) - 8 cos^{2} (x) = 0

用替代法求解

4 - 4 cos (x) - 8 cos^{2} (x) = 0

令

: cos (x) = u

4 - 4 u - 8 u^{2} = 0

4 - 4 u - 8 u^{2} = 0 : u = - 1, u = \frac{1}{2}

4 - 4 u - 8 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 8, b = - 4, c = 4

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

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

(- 4)^{2} - 4 (- 8) \cdot 4 = 12

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

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

= 4^{2} + 4 \cdot 8 \cdot 4

数字相乘:

4 \cdot 8 \cdot 4 = 128

= 4^{2} + 128

4^{2} = 16

= 16 + 128

数字相加:

16 + 128 = 144

= 144

因式分解数字:

144 = 1 2^{2}

= 1 2^{2}

使用根式运算法则:

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- ( - 4 ) \pm 12}{2 ( - 8 )}

将解分隔开

u_{1} = \frac{- ( - 4 ) + 12}{2 ( - 8 )}, u_{2} = \frac{- ( - 4 ) - 12}{2 ( - 8 )}

u = \frac{- ( - 4 ) + 12}{2 ( - 8 )} : - 1

\frac{- ( - 4 ) + 12}{2 ( - 8 )}

去除括号:

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

= \frac{4 + 12}{- 2 \cdot 8}

数字相加:

4 + 12 = 16

= \frac{16}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{16}{- 16}

使用分式法则:

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

= - \frac{16}{16}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 4 ) - 12}{2 ( - 8 )} : \frac{1}{2}

\frac{- ( - 4 ) - 12}{2 ( - 8 )}

去除括号:

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

= \frac{4 - 12}{- 2 \cdot 8}

数字相减:

4 - 12 = - 8

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

数字相乘:

2 \cdot 8 = 16

= \frac{- 8}{- 16}

使用分式法则:

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

= \frac{8}{16}

约分:

8

= \frac{1}{2}

二次方程组的解是:

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

u = cos (x)

代回

cos (x) = - 1, cos (x) = \frac{1}{2}

cos (x) = - 1, cos (x) = \frac{1}{2}

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

cos (x) = - 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}

x = π + 2 πn

x = π + 2 πn

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}

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}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

合并所有解

x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

π + 2 πn

的解:

真

π + 2 πn

代入

n = 1

π + 2 π 1

对于

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

代入

x = π + 2 π 1

6 sin (π + 2 π 1) - 2 cos (π + 2 π 1) = 2

整理后得

1.41421 \dots = 1.41421 \dots

\Rightarrow 真

检验

\frac{π}{3} + 2 πn

的解:

真

\frac{π}{3} + 2 πn

代入

n = 1

\frac{π}{3} + 2 π 1

对于

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

代入

x = \frac{π}{3} + 2 π 1

6 sin (\frac{π}{3} + 2 π 1) - 2 cos (\frac{π}{3} + 2 π 1) = 2

整理后得

1.41421 \dots = 1.41421 \dots

\Rightarrow 真

检验

\frac{5 π}{3} + 2 πn

的解:

假

\frac{5 π}{3} + 2 πn

代入

n = 1

\frac{5 π}{3} + 2 π 1

对于

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

代入

x = \frac{5 π}{3} + 2 π 1

6 sin (\frac{5 π}{3} + 2 π 1) - 2 cos (\frac{5 π}{3} + 2 π 1) = 2

整理后得

- 2.82842 \dots = 1.41421 \dots

\Rightarrow 假

x = π + 2 πn, x = \frac{π}{3} + 2 πn

作图

流行的例子

3-3sin(x)=6cos^2(x)

cos(x)= 60/61 ,tan(x)= 11/60

3sqrt(2)cos(x)+2=-1

4=tan^2(45+x/2)