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受欢迎的三角函数 >

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

解答

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

解答

x = 2 πn

+1

度数

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

求解步骤

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

使用三角恒等式改写

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

化简

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

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

用替代法求解

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

令

: cos (x) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 9)^{2} - 4 (- 1) \cdot 10 = 11

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

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

= 9^{2} + 4 \cdot 1 \cdot 10

数字相乘:

4 \cdot 1 \cdot 10 = 40

= 9^{2} + 40

9^{2} = 81

= 81 + 40

数字相加:

81 + 40 = 121

= 121

因式分解数字:

121 = 1 1^{2}

= 1 1^{2}

使用根式运算法则:

n a^{n} = a

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- ( - 9 ) \pm 11}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- ( - 9 ) + 11}{2 ( - 1 )}, u_{2} = \frac{- ( - 9 ) - 11}{2 ( - 1 )}

u = \frac{- ( - 9 ) + 11}{2 ( - 1 )} : - 10

\frac{- ( - 9 ) + 11}{2 ( - 1 )}

去除括号:

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

= \frac{9 + 11}{- 2 \cdot 1}

数字相加:

9 + 11 = 20

= \frac{20}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{20}{- 2}

使用分式法则:

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

= - \frac{20}{2}

数字相除:

\frac{20}{2} = 10

= - 10

u = \frac{- ( - 9 ) - 11}{2 ( - 1 )} : 1

\frac{- ( - 9 ) - 11}{2 ( - 1 )}

去除括号:

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

= \frac{9 - 11}{- 2 \cdot 1}

数字相减:

9 - 11 = - 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 = - 10, u = 1

u = cos (x)

代回

cos (x) = - 10, cos (x) = 1

cos (x) = - 10, cos (x) = 1

cos (x) = - 10 :

无解

cos (x) = - 10

- 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

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cos (x) = 0, 72