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受欢迎的三角函数 >

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

解答

cos (x) = 1 + sin (x)

解答

x = \frac{3 π}{2} + 2 πn, x = 2 πn

+1

度数

x = 27 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

求解步骤

cos (x) = 1 + sin (x)

两边进行平方

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

两边减去

1 + sin (x)^{2}

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

用替代法求解

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

令

: sin (x) = u

- u - u^{2} = 0

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

- u - u^{2} = 0

改写成标准形式

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

- u^{2} - u = 0

使用求根公式求解

- u^{2} - u = 0

二次方程求根公式:

若

a = - 1, b = - 1, c = 0

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

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

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

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 1 + 0

数字相加:

1 + 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

\frac{- ( - 1 ) + 1}{2 ( - 1 )}

去除括号:

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

= \frac{1 + 1}{- 2 \cdot 1}

数字相加:

1 + 1 = 2

= \frac{2}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2}{- 2}

使用分式法则:

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

= - \frac{2}{2}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 1 ) - 1}{2 ( - 1 )} : 0

\frac{- ( - 1 ) - 1}{2 ( - 1 )}

去除括号:

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

= \frac{1 - 1}{- 2 \cdot 1}

数字相减:

1 - 1 = 0

= \frac{0}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{- 2}

使用分式法则:

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

= - \frac{0}{2}

使用法则

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

= - 0

= 0

二次方程组的解是:

u = - 1, u = 0

u = sin (x)

代回

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

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

sin (x) = - 1 : x = \frac{3 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{3 π}{2} + 2 πn

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

cos (x) = 1 + sin (x)

检验解是否符合

去除与方程不符的解。

检验

\frac{3 π}{2} + 2 πn

的解:

真

\frac{3 π}{2} + 2 πn

代入

n = 1

\frac{3 π}{2} + 2 π 1

对于

cos (x) = 1 + sin (x)

代入

x = \frac{3 π}{2} + 2 π 1

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

整理后得

0 = 0

\Rightarrow 真

检验

2 πn

的解:

真

2 πn

代入

n = 1

2 π 1

对于

cos (x) = 1 + sin (x)

代入

x = 2 π 1

cos (2 π 1) = 1 + sin (2 π 1)

整理后得

1 = 1

\Rightarrow 真

检验

π + 2 πn

的解:

假

π + 2 πn

代入

n = 1

π + 2 π 1

对于

cos (x) = 1 + sin (x)

代入

x = π + 2 π 1

cos (π + 2 π 1) = 1 + sin (π + 2 π 1)

整理后得

- 1 = 1

\Rightarrow 假

x = \frac{3 π}{2} + 2 πn, x = 2 πn

作图

流行的例子

cot (x - \frac{π}{4}) = - 1

3 cos (θ) + 1 = 1

sin (θ) = 3

sqrt(2)cos(x)csc(x)=2cos(x)

2 cos (x) csc (x) = 2 cos (x)

-2cos^2(x)=3cos(x)+1

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