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受欢迎的三角函数 >

sqrt(3)cos(x)+sin(x)=1

解答

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

解答

x = \frac{11 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

+1

度数

x = 33 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

sin (x)

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

两边进行平方

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

两边减去

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

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

使用三角恒等式改写

- 1 - sin^{2} (x) + 2 sin (x) + 3 cos^{2} (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)

= - 1 - sin^{2} (x) + 2 sin (x) + 3 (1 - sin^{2} (x))

化简

- 1 - sin^{2} (x) + 2 sin (x) + 3 (1 - sin^{2} (x)) : 2 sin (x) - 4 sin^{2} (x) + 2

- 1 - sin^{2} (x) + 2 sin (x) + 3 (1 - sin^{2} (x))

乘开

3 (1 - sin^{2} (x)) : 3 - 3 sin^{2} (x)

3 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (x)

= 3 \cdot 1 - 3 sin^{2} (x)

数字相乘:

3 \cdot 1 = 3

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

= - 1 - sin^{2} (x) + 2 sin (x) + 3 - 3 sin^{2} (x)

化简

- 1 - sin^{2} (x) + 2 sin (x) + 3 - 3 sin^{2} (x) : 2 sin (x) - 4 sin^{2} (x) + 2

- 1 - sin^{2} (x) + 2 sin (x) + 3 - 3 sin^{2} (x)

对同类项分组

= - sin^{2} (x) + 2 sin (x) - 3 sin^{2} (x) - 1 + 3

同类项相加:

- sin^{2} (x) - 3 sin^{2} (x) = - 4 sin^{2} (x)

= - 4 sin^{2} (x) + 2 sin (x) - 1 + 3

数字相加/相减:

- 1 + 3 = 2

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

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

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

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

用替代法求解

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

令

: sin (x) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

2^{2} - 4 (- 4) \cdot 2 = 6

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 4 \cdot 2 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

数字相加:

4 + 32 = 36

= 36

因式分解数字:

36 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a

6^{2} = 6

= 6

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

将解分隔开

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

u = \frac{- 2 + 6}{2 ( - 4 )} : - \frac{1}{2}

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

去除括号:

(- a) = - a

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

数字相加/相减:

- 2 + 6 = 4

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

数字相乘:

2 \cdot 4 = 8

= \frac{4}{- 8}

使用分式法则:

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

= - \frac{4}{8}

约分:

4

= - \frac{1}{2}

u = \frac{- 2 - 6}{2 ( - 4 )} : 1

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

去除括号:

(- a) = - a

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

数字相减:

- 2 - 6 = - 8

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 8}{- 8}

使用分式法则:

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

= \frac{8}{8}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

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

u = sin (x)

代回

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

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

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

sin (x) = - \frac{1}{2}

sin (x) = - \frac{1}{2}

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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 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 = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

\frac{7 π}{6} + 2 πn

的解:

假

\frac{7 π}{6} + 2 πn

代入

n = 1

\frac{7 π}{6} + 2 π 1

对于

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

代入

x = \frac{7 π}{6} + 2 π 1

3 cos (\frac{7 π}{6} + 2 π 1) + sin (\frac{7 π}{6} + 2 π 1) = 1

整理后得

- 2 = 1

\Rightarrow 假

检验

\frac{11 π}{6} + 2 πn

的解:

真

\frac{11 π}{6} + 2 πn

代入

n = 1

\frac{11 π}{6} + 2 π 1

对于

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

代入

x = \frac{11 π}{6} + 2 π 1

3 cos (\frac{11 π}{6} + 2 π 1) + sin (\frac{11 π}{6} + 2 π 1) = 1

整理后得

1 = 1

\Rightarrow 真

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

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

代入

x = \frac{π}{2} + 2 π 1

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

整理后得

1 = 1

\Rightarrow 真

x = \frac{11 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

作图

流行的例子

sin(2x+15)=cos(1/2 x-15)

sin (2 x + 1 5^{\circ}) = cos (\frac{1}{2} x - 1 5^{\circ})

sin (x) = 0.37

6 sin (x) + 3 = 0

2cos(x)=2cos(3x)

2 cos (x) = 2 cos (3 x)

8cos^2(x)-2cos(x)-1=0

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