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受欢迎的三角函数 >

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

解答

1 + cos (x) = sin (x)

解答

x = \frac{π}{2} + 2 πn, x = π + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

1 + cos (x) = sin (x)

两边进行平方

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

两边减去

sin^{2} (x)

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

cos (x) + cos^{2} (x) = 0

用替代法求解

cos (x) + cos^{2} (x) = 0

令

: cos (x) = u

u + u^{2} = 0

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

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 \cdot 1 \cdot 0}{2 \cdot 1}

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 1 - 0

数字相减:

1 - 0 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- 1 \pm 1}{2 \cdot 1}

将解分隔开

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

u = \frac{- 1 + 1}{2 \cdot 1} : 0

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

数字相加/相减:

- 1 + 1 = 0

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

数字相乘:

2 \cdot 1 = 2

= \frac{0}{2}

使用法则

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

= 0

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

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

数字相减:

- 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

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 = π + 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 = π + 2 πn

x = π + 2 πn

合并所有解

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

将解代入原方程进行验证

将它们代入

1 + cos (x) = sin (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

1 + cos (x) = sin (x)

代入

x = \frac{π}{2} + 2 π 1

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

1 + cos (x) = sin (x)

代入

x = \frac{3 π}{2} + 2 π 1

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

整理后得

1 = - 1

\Rightarrow 假

检验

π + 2 πn

的解:

真

π + 2 πn

代入

n = 1

π + 2 π 1

对于

1 + cos (x) = sin (x)

代入

x = π + 2 π 1

1 + cos (π + 2 π 1) = sin (π + 2 π 1)

整理后得

0 = 0

\Rightarrow 真

x = \frac{π}{2} + 2 πn, x = π + 2 πn

作图

流行的例子

6 sin (3 t) = 0

sin(3x)cos(6x)-cos(3x)sin(6x)=-0.9

sin (3 x) cos (6 x) - cos (3 x) sin (6 x) = - 0.9

0 = cos (t)

∣ sin (x) ∣ = 1

tan (x) csc (x) = 2