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受欢迎的三角函数 >

10(1-cos(x))=sin^2(x)

解答

10 (1 - cos (x)) = sin^{2} (x)

解答

x = 2 πn

+1

度数

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

求解步骤

10 (1 - cos (x)) = sin^{2} (x)

两边减去

sin^{2} (x)

10 (1 - cos (x)) - sin^{2} (x) = 0

使用三角恒等式改写

- sin^{2} (x) + (1 - cos (x)) \cdot 10

使用毕达哥拉斯恒等式:

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

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

= - (1 - cos^{2} (x)) + (1 - cos (x)) \cdot 10

化简

- (1 - cos^{2} (x)) + (1 - cos (x)) \cdot 10 : cos^{2} (x) - 10 cos (x) + 9

- (1 - cos^{2} (x)) + (1 - cos (x)) \cdot 10

= - (1 - cos^{2} (x)) + 10 (1 - cos (x))

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

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

打开括号

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

使用加减运算法则

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

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

= - 1 + cos^{2} (x) + (1 - cos (x)) \cdot 10

乘开

10 (1 - cos (x)) : 10 - 10 cos (x)

10 (1 - cos (x))

使用分配律:

a (b - c) = ab - a c

a = 10, b = 1, c = cos (x)

= 10 \cdot 1 - 10 cos (x)

数字相乘:

10 \cdot 1 = 10

= 10 - 10 cos (x)

= - 1 + cos^{2} (x) + 10 - 10 cos (x)

化简

- 1 + cos^{2} (x) + 10 - 10 cos (x) : cos^{2} (x) - 10 cos (x) + 9

- 1 + cos^{2} (x) + 10 - 10 cos (x)

对同类项分组

= cos^{2} (x) - 10 cos (x) - 1 + 10

数字相加/相减:

- 1 + 10 = 9

= cos^{2} (x) - 10 cos (x) + 9

= cos^{2} (x) - 10 cos (x) + 9

= cos^{2} (x) - 10 cos (x) + 9

9 + cos^{2} (x) - 10 cos (x) = 0

用替代法求解

9 + cos^{2} (x) - 10 cos (x) = 0

令

: cos (x) = u

9 + u^{2} - 10 u = 0

9 + u^{2} - 10 u = 0 : u = 9, u = 1

9 + u^{2} - 10 u = 0

改写成标准形式

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

u^{2} - 10 u + 9 = 0

使用求根公式求解

u^{2} - 10 u + 9 = 0

二次方程求根公式:

若

a = 1, b = - 10, c = 9

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 9}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 9}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot 9 = 8

(- 10)^{2} - 4 \cdot 1 \cdot 9

使用指数法则:

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

若

n

是偶数

(- 10)^{2} = 1 0^{2}

= 1 0^{2} - 4 \cdot 1 \cdot 9

数字相乘:

4 \cdot 1 \cdot 9 = 36

= 1 0^{2} - 36

1 0^{2} = 100

= 100 - 36

数字相减:

100 - 36 = 64

= 64

因式分解数字:

64 = 8^{2}

= 8^{2}

使用根式运算法则:

n a^{n} = a

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 10 ) \pm 8}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 10 ) + 8}{2 \cdot 1}, u_{2} = \frac{- ( - 10 ) - 8}{2 \cdot 1}

u = \frac{- ( - 10 ) + 8}{2 \cdot 1} : 9

\frac{- ( - 10 ) + 8}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{10 + 8}{2 \cdot 1}

数字相加:

10 + 8 = 18

= \frac{18}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{18}{2}

数字相除:

\frac{18}{2} = 9

= 9

u = \frac{- ( - 10 ) - 8}{2 \cdot 1} : 1

\frac{- ( - 10 ) - 8}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{10 - 8}{2 \cdot 1}

数字相减:

10 - 8 = 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = 9, u = 1

u = cos (x)

代回

cos (x) = 9, cos (x) = 1

cos (x) = 9, cos (x) = 1

cos (x) = 9 :

无解

cos (x) = 9

- 1 \leq cos (x) \leq 1

无解

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 = 2 πn

作图

流行的例子

2 sin (4 x) = 0

9*csc(x)=6sqrt(3),(-infinity ,infinity)

9 \cdot csc (x) = 63, (- \infty, \infty)

cos (2 x - \frac{π}{2}) = 1

tan(x)=-sqrt(2)

tan (x) = - 2

sin (θ) = \frac{1}{8}