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受欢迎的三角函数 >

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

解答

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

解答

x = \frac{π}{2} + 2 πn, x = 2 πn

+1

度数

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

求解步骤

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

两边进行平方

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

两边减去

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

9 sin^{2} (x) - 9 + 18 cos (x) - 9 cos^{2} (x) = 0

使用三角恒等式改写

- 9 + 18 cos (x) - 9 cos^{2} (x) + 9 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)

= - 9 + 18 cos (x) - 9 cos^{2} (x) + 9 (1 - cos^{2} (x))

化简

- 9 + 18 cos (x) - 9 cos^{2} (x) + 9 (1 - cos^{2} (x)) : 18 cos (x) - 18 cos^{2} (x)

- 9 + 18 cos (x) - 9 cos^{2} (x) + 9 (1 - cos^{2} (x))

乘开

9 (1 - cos^{2} (x)) : 9 - 9 cos^{2} (x)

9 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 9, b = 1, c = cos^{2} (x)

= 9 \cdot 1 - 9 cos^{2} (x)

数字相乘:

9 \cdot 1 = 9

= 9 - 9 cos^{2} (x)

= - 9 + 18 cos (x) - 9 cos^{2} (x) + 9 - 9 cos^{2} (x)

化简

- 9 + 18 cos (x) - 9 cos^{2} (x) + 9 - 9 cos^{2} (x) : 18 cos (x) - 18 cos^{2} (x)

- 9 + 18 cos (x) - 9 cos^{2} (x) + 9 - 9 cos^{2} (x)

对同类项分组

= 18 cos (x) - 9 cos^{2} (x) - 9 cos^{2} (x) - 9 + 9

同类项相加:

- 9 cos^{2} (x) - 9 cos^{2} (x) = - 18 cos^{2} (x)

= 18 cos (x) - 18 cos^{2} (x) - 9 + 9

- 9 + 9 = 0

= 18 cos (x) - 18 cos^{2} (x)

= 18 cos (x) - 18 cos^{2} (x)

= 18 cos (x) - 18 cos^{2} (x)

18 cos (x) - 18 cos^{2} (x) = 0

用替代法求解

18 cos (x) - 18 cos^{2} (x) = 0

令

: cos (x) = u

18 u - 18 u^{2} = 0

18 u - 18 u^{2} = 0 : u = 0, u = 1

18 u - 18 u^{2} = 0

改写成标准形式

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

- 18 u^{2} + 18 u = 0

使用求根公式求解

- 18 u^{2} + 18 u = 0

二次方程求根公式:

若

a = - 18, b = 18, c = 0

u_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 ( - 18 ) \cdot 0}{2 ( - 18 )}

u_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 ( - 18 ) \cdot 0}{2 ( - 18 )}

1 8^{2} - 4 (- 18) \cdot 0 = 18

1 8^{2} - 4 (- 18) \cdot 0

使用法则

- (- a) = a

= 1 8^{2} + 4 \cdot 18 \cdot 0

使用法则

0 \cdot a = 0

= 1 8^{2} + 0

1 8^{2} + 0 = 1 8^{2}

= 1 8^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 18

u_{1, 2} = \frac{- 18 \pm 18}{2 ( - 18 )}

将解分隔开

u_{1} = \frac{- 18 + 18}{2 ( - 18 )}, u_{2} = \frac{- 18 - 18}{2 ( - 18 )}

u = \frac{- 18 + 18}{2 ( - 18 )} : 0

\frac{- 18 + 18}{2 ( - 18 )}

去除括号:

(- a) = - a

= \frac{- 18 + 18}{- 2 \cdot 18}

数字相加/相减:

- 18 + 18 = 0

= \frac{0}{- 2 \cdot 18}

数字相乘:

2 \cdot 18 = 36

= \frac{0}{- 36}

使用分式法则:

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

= - \frac{0}{36}

使用法则

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

= - 0

= 0

u = \frac{- 18 - 18}{2 ( - 18 )} : 1

\frac{- 18 - 18}{2 ( - 18 )}

去除括号:

(- a) = - a

= \frac{- 18 - 18}{- 2 \cdot 18}

数字相减:

- 18 - 18 = - 36

= \frac{- 36}{- 2 \cdot 18}

数字相乘:

2 \cdot 18 = 36

= \frac{- 36}{- 36}

使用分式法则:

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

= \frac{36}{36}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = 0, u = 1

u = cos (x)

代回

cos (x) = 0, cos (x) = 1

cos (x) = 0, cos (x) = 1

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) = 1 : x = 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 = 0 + 2 πn

x = 0 + 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

合并所有解

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

将解代入原方程进行验证

将它们代入

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

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

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

代入

x = \frac{π}{2} + 2 π 1

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

整理后得

3 = 3

\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

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

代入

x = \frac{3 π}{2} + 2 π 1

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

整理后得

- 3 = 3

\Rightarrow 假

检验

2 πn

的解:

真

2 πn

代入

n = 1

2 π 1

对于

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

代入

x = 2 π 1

3 sin (2 π 1) = 3 - 3 cos (2 π 1)

整理后得

0 = 0

\Rightarrow 真

x = \frac{π}{2} + 2 πn, x = 2 πn

作图

流行的例子

0 = 1 - 2 cos (x)

-5sin^2(θ)+17cos(θ)+9=4cos(θ)-2

- 5 sin^{2} (θ) + 17 cos (θ) + 9 = 4 cos (θ) - 2

(cos(3θ))(cos(θ))+1=(sin(3θ))(sin(θ))

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

4 sin (x) - 3 = 0

9sin(x)tan(x)=2tan(x)

9 sin (x) tan (x) = 2 tan (x)