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受欢迎的三角函数 >

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

解答

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

解答

x = 2 πn, x = \frac{π}{2} + 2 πn

+1

度数

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

求解步骤

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

两边减去

2 sin (x)

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

两边进行平方

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

两边减去

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

= - 4 + 4 (1 - sin^{2} (x)) - 4 sin^{2} (x) + 8 sin (x)

化简

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

- 4 + 4 (1 - sin^{2} (x)) - 4 sin^{2} (x) + 8 sin (x)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

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

数字相乘:

4 \cdot 1 = 4

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

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

化简

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

- 4 + 4 - 4 sin^{2} (x) - 4 sin^{2} (x) + 8 sin (x)

同类项相加:

- 4 sin^{2} (x) - 4 sin^{2} (x) = - 8 sin^{2} (x)

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

- 4 + 4 = 0

= 8 sin (x) - 8 sin^{2} (x)

= 8 sin (x) - 8 sin^{2} (x)

= 8 sin (x) - 8 sin^{2} (x)

8 sin (x) - 8 sin^{2} (x) = 0

用替代法求解

8 sin (x) - 8 sin^{2} (x) = 0

令

: sin (x) = u

8 u - 8 u^{2} = 0

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

8 u - 8 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

= 8^{2} + 4 \cdot 8 \cdot 0

使用法则

0 \cdot a = 0

= 8^{2} + 0

8^{2} + 0 = 8^{2}

= 8^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 8

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

将解分隔开

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

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

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

去除括号:

(- a) = - a

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

数字相加/相减:

- 8 + 8 = 0

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

数字相乘:

2 \cdot 8 = 16

= \frac{0}{- 16}

使用分式法则:

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

= - \frac{0}{16}

使用法则

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

= - 0

= 0

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

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

去除括号:

(- a) = - a

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

数字相减:

- 8 - 8 = - 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

二次方程组的解是:

u = 0, u = 1

u = sin (x)

代回

sin (x) = 0, sin (x) = 1

sin (x) = 0, sin (x) = 1

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 0

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

sin (x) = 1

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

合并所有解

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

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

2 πn

的解:

真

2 πn

代入

n = 1

2 π 1

对于

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

代入

x = 2 π 1

2 cos (2 π 1) + 2 sin (2 π 1) = 2

整理后得

2 = 2

\Rightarrow 真

检验

π + 2 πn

的解:

假

π + 2 πn

代入

n = 1

π + 2 π 1

对于

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

代入

x = π + 2 π 1

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

整理后得

- 2 = 2

\Rightarrow 假

检验

\frac{π}{2} + 2 πn

的解:

真

\frac{π}{2} + 2 πn

代入

n = 1

\frac{π}{2} + 2 π 1

对于

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

代入

x = \frac{π}{2} + 2 π 1

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

整理后得

2 = 2

\Rightarrow 真

x = 2 πn, x = \frac{π}{2} + 2 πn

作图

流行的例子

cos(3x)=cos(6x)

cos (3 x) = cos (6 x)

cos (A) = \frac{5}{9}

cos (x) = 0.56

3 - cot^{2} (x) = 0

(-4cos(x)+2sin(x))^2+12sin^2(x)=24

(- 4 cos (x) + 2 sin (x))^{2} + 12 sin^{2} (x) = 24