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受欢迎的三角函数 >

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

解答

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

解答

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

+1

度数

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

求解步骤

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

两边减去

3 cos (x)

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

两边进行平方

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

两边减去

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

sin^{2} (x) - 1 + 23 cos (x) - 3 cos^{2} (x) = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

= - 3 cos^{2} (x) + 23 cos (x) - cos^{2} (x)

化简

= - 4 cos^{2} (x) + 23 cos (x)

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

用替代法求解

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

令

: cos (x) = u

- 4 u^{2} + 2 u 3 = 0

- 4 u^{2} + 2 u 3 = 0 : u = 0, u = \frac{3}{2}

- 4 u^{2} + 2 u 3 = 0

改写成标准形式

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

- 4 u^{2} + 23 u = 0

使用求根公式求解

- 4 u^{2} + 23 u = 0

二次方程求根公式:

若

a = - 4, b = 23, c = 0

u_{1, 2} = \frac{- 2 3 \pm ( 2 3 ) ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

u_{1, 2} = \frac{- 2 3 \pm ( 2 3 ) ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

(23)^{2} - 4 (- 4) \cdot 0 = 23

(23)^{2} - 4 (- 4) \cdot 0

使用法则

- (- a) = a

= (23)^{2} + 4 \cdot 4 \cdot 0

(23)^{2} = 2^{2} \cdot 3

(23)^{2}

使用指数法则:

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

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

(3)^{2} : 3

使用根式运算法则:

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

= (3^{\frac{1}{2}})^{2}

使用指数法则:

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

= 3^{\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

= 3

= 2^{2} \cdot 3

4 \cdot 4 \cdot 0 = 0

4 \cdot 4 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 2^{2} \cdot 3 + 0

2^{2} \cdot 3 + 0 = 2^{2} \cdot 3

= 2^{2} \cdot 3

使用根式运算法则:

n ab = n a n b,

假定

a \geq 0, b \geq 0

= 3 2^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

2^{2} = 2

= 23

u_{1, 2} = \frac{- 2 3 \pm 2 3}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- 2 3 + 2 3}{2 ( - 4 )}, u_{2} = \frac{- 2 3 - 2 3}{2 ( - 4 )}

u = \frac{- 2 3 + 2 3}{2 ( - 4 )} : 0

\frac{- 2 3 + 2 3}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 2 3 + 2 3}{- 2 \cdot 4}

同类项相加:

- 23 + 23 = 0

= \frac{0}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{0}{- 8}

使用分式法则:

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

= - \frac{0}{8}

使用法则

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

= - 0

= 0

u = \frac{- 2 3 - 2 3}{2 ( - 4 )} : \frac{3}{2}

\frac{- 2 3 - 2 3}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 2 3 - 2 3}{- 2 \cdot 4}

同类项相加:

- 23 - 23 = - 43

= \frac{- 4 3}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 4 3}{- 8}

使用分式法则:

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

= \frac{4 3}{8}

约分:

4

= \frac{3}{2}

二次方程组的解是:

u = 0, u = \frac{3}{2}

u = cos (x)

代回

cos (x) = 0, cos (x) = \frac{3}{2}

cos (x) = 0, cos (x) = \frac{3}{2}

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0

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}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

cos (x) = \frac{3}{2} : x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

cos (x) = \frac{3}{2}

cos (x) = \frac{3}{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}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

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

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

代入

x = \frac{π}{2} + 2 π 1

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

整理后得

1 = 1

\Rightarrow 真

检验

\frac{3 π}{2} + 2 πn

的解:

假

\frac{3 π}{2} + 2 πn

代入

n = 1

\frac{3 π}{2} + 2 π 1

对于

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

代入

x = \frac{3 π}{2} + 2 π 1

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

整理后得

- 1 = 1

\Rightarrow 假

检验

\frac{π}{6} + 2 πn

的解:

假

\frac{π}{6} + 2 πn

代入

n = 1

\frac{π}{6} + 2 π 1

对于

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

代入

x = \frac{π}{6} + 2 π 1

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

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

代入

x = \frac{11 π}{6} + 2 π 1

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

整理后得

1 = 1

\Rightarrow 真

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

作图

流行的例子

cos(x)+sin(2x)=0

cos (x) + sin (2 x) = 0

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

cos (θ) = \frac{5}{13}

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

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

sin^2(x)=cos(x)

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