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受欢迎的三角函数 >

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

解答

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

解答

x = - 0.92729 \dots + 2 πn, x = 2 πn

+1

度数

x = - 53.13010 \dots^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边进行平方

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

两边减去

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

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

使用三角恒等式改写

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

化简

- 4 - sin^{2} (x) + 4 (1 - sin^{2} (x)) - 4 sin (x) : - 5 sin^{2} (x) - 4 sin (x)

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

化简

- 4 - sin^{2} (x) + 4 - 4 sin^{2} (x) - 4 sin (x) : - 5 sin^{2} (x) - 4 sin (x)

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

对同类项分组

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

同类项相加:

- sin^{2} (x) - 4 sin^{2} (x) = - 5 sin^{2} (x)

= - 5 sin^{2} (x) - 4 sin (x) - 4 + 4

- 4 + 4 = 0

= - 5 sin^{2} (x) - 4 sin (x)

= - 5 sin^{2} (x) - 4 sin (x)

= - 5 sin^{2} (x) - 4 sin (x)

- 4 sin (x) - 5 sin^{2} (x) = 0

用替代法求解

- 4 sin (x) - 5 sin^{2} (x) = 0

令

: sin (x) = u

- 4 u - 5 u^{2} = 0

- 4 u - 5 u^{2} = 0 : u = - \frac{4}{5}, u = 0

- 4 u - 5 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 5, b = - 4, c = 0

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 5 ) \cdot 0}{2 ( - 5 )}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 5 ) \cdot 0}{2 ( - 5 )}

(- 4)^{2} - 4 (- 5) \cdot 0 = 4

(- 4)^{2} - 4 (- 5) \cdot 0

使用法则

- (- a) = a

= (- 4)^{2} + 4 \cdot 5 \cdot 0

使用指数法则:

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

若

n

是偶数

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

= 4^{2} + 4 \cdot 5 \cdot 0

使用法则

0 \cdot a = 0

= 4^{2} + 0

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

= 4^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 4

u_{1, 2} = \frac{- ( - 4 ) \pm 4}{2 ( - 5 )}

将解分隔开

u_{1} = \frac{- ( - 4 ) + 4}{2 ( - 5 )}, u_{2} = \frac{- ( - 4 ) - 4}{2 ( - 5 )}

u = \frac{- ( - 4 ) + 4}{2 ( - 5 )} : - \frac{4}{5}

\frac{- ( - 4 ) + 4}{2 ( - 5 )}

去除括号:

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

= \frac{4 + 4}{- 2 \cdot 5}

数字相加:

4 + 4 = 8

= \frac{8}{- 2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{8}{- 10}

使用分式法则:

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

= - \frac{8}{10}

约分:

2

= - \frac{4}{5}

u = \frac{- ( - 4 ) - 4}{2 ( - 5 )} : 0

\frac{- ( - 4 ) - 4}{2 ( - 5 )}

去除括号:

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

= \frac{4 - 4}{- 2 \cdot 5}

数字相减:

4 - 4 = 0

= \frac{0}{- 2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{0}{- 10}

使用分式法则:

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

= - \frac{0}{10}

使用法则

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

= - 0

= 0

二次方程组的解是:

u = - \frac{4}{5}, u = 0

u = sin (x)

代回

sin (x) = - \frac{4}{5}, sin (x) = 0

sin (x) = - \frac{4}{5}, sin (x) = 0

sin (x) = - \frac{4}{5} : x = arcsin (- \frac{4}{5}) + 2 πn, x = π + arcsin (\frac{4}{5}) + 2 πn

sin (x) = - \frac{4}{5}

使用反三角函数性质

sin (x) = - \frac{4}{5}

sin (x) = - \frac{4}{5}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{4}{5}) + 2 πn, x = π + arcsin (\frac{4}{5}) + 2 πn

x = arcsin (- \frac{4}{5}) + 2 πn, x = π + arcsin (\frac{4}{5}) + 2 πn

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

合并所有解

x = arcsin (- \frac{4}{5}) + 2 πn, x = π + arcsin (\frac{4}{5}) + 2 πn, x = 2 πn, x = π + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

arcsin (- \frac{4}{5}) + 2 πn

的解:

真

arcsin (- \frac{4}{5}) + 2 πn

代入

n = 1

arcsin (- \frac{4}{5}) + 2 π 1

对于

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

代入

x = arcsin (- \frac{4}{5}) + 2 π 1

2 cos (arcsin (- \frac{4}{5}) + 2 π 1) = 2 + sin (arcsin (- \frac{4}{5}) + 2 π 1)

整理后得

1.2 = 1.2

\Rightarrow 真

检验

π + arcsin (\frac{4}{5}) + 2 πn

的解:

假

π + arcsin (\frac{4}{5}) + 2 πn

代入

n = 1

π + arcsin (\frac{4}{5}) + 2 π 1

对于

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

代入

x = π + arcsin (\frac{4}{5}) + 2 π 1

2 cos (π + arcsin (\frac{4}{5}) + 2 π 1) = 2 + sin (π + arcsin (\frac{4}{5}) + 2 π 1)

整理后得

- 1.2 = 1.2

\Rightarrow 假

检验

2 πn

的解:

真

2 πn

代入

n = 1

2 π 1

对于

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

代入

x = 2 π 1

2 cos (2 π 1) = 2 + sin (2 π 1)

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)

代入

x = π + 2 π 1

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

整理后得

- 2 = 2

\Rightarrow 假

x = arcsin (- \frac{4}{5}) + 2 πn, x = 2 πn

以小数形式表示解

x = - 0.92729 \dots + 2 πn, x = 2 πn

作图

流行的例子

3sqrt(2)cos(θ)+1=-2

32 cos (θ) + 1 = - 2

5sin(2θ)=9tan(θ)

5 sin (2 θ) = 9 tan (θ)

0 = 1 - 2 sin (x)

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

cos (\frac{θ}{2}) = - \frac{1}{2}