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受欢迎的三角函数 >

8sin^2(x)-10cos(x)-11=0

解答

8 sin^{2} (x) - 10 cos (x) - 11 = 0

解答

x = 2.41885 \dots + 2 πn, x = - 2.41885 \dots + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

+1

度数

x = 138.59037 \dots^{\circ} + 36 0^{\circ} n, x = - 138.59037 \dots^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

求解步骤

8 sin^{2} (x) - 10 cos (x) - 11 = 0

使用三角恒等式改写

- 11 - 10 cos (x) + 8 sin^{2} (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)

= - 11 - 10 cos (x) + 8 (1 - cos^{2} (x))

化简

- 11 - 10 cos (x) + 8 (1 - cos^{2} (x)) : - 8 cos^{2} (x) - 10 cos (x) - 3

- 11 - 10 cos (x) + 8 (1 - cos^{2} (x))

乘开

8 (1 - cos^{2} (x)) : 8 - 8 cos^{2} (x)

8 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 8, b = 1, c = cos^{2} (x)

= 8 \cdot 1 - 8 cos^{2} (x)

数字相乘:

8 \cdot 1 = 8

= 8 - 8 cos^{2} (x)

= - 11 - 10 cos (x) + 8 - 8 cos^{2} (x)

化简

- 11 - 10 cos (x) + 8 - 8 cos^{2} (x) : - 8 cos^{2} (x) - 10 cos (x) - 3

- 11 - 10 cos (x) + 8 - 8 cos^{2} (x)

对同类项分组

= - 10 cos (x) - 8 cos^{2} (x) - 11 + 8

数字相加/相减:

- 11 + 8 = - 3

= - 8 cos^{2} (x) - 10 cos (x) - 3

= - 8 cos^{2} (x) - 10 cos (x) - 3

= - 8 cos^{2} (x) - 10 cos (x) - 3

- 3 - 10 cos (x) - 8 cos^{2} (x) = 0

用替代法求解

- 3 - 10 cos (x) - 8 cos^{2} (x) = 0

令

: cos (x) = u

- 3 - 10 u - 8 u^{2} = 0

- 3 - 10 u - 8 u^{2} = 0 : u = - \frac{3}{4}, u = - \frac{1}{2}

- 3 - 10 u - 8 u^{2} = 0

改写成标准形式

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

- 8 u^{2} - 10 u - 3 = 0

使用求根公式求解

- 8 u^{2} - 10 u - 3 = 0

二次方程求根公式:

若

a = - 8, b = - 10, c = - 3

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 ( - 8 ) ( - 3 )}{2 ( - 8 )}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 ( - 8 ) ( - 3 )}{2 ( - 8 )}

(- 10)^{2} - 4 (- 8) (- 3) = 2

(- 10)^{2} - 4 (- 8) (- 3)

使用法则

- (- a) = a

= (- 10)^{2} - 4 \cdot 8 \cdot 3

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 8 \cdot 3

数字相乘:

4 \cdot 8 \cdot 3 = 96

= 1 0^{2} - 96

1 0^{2} = 100

= 100 - 96

数字相减:

100 - 96 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

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

将解分隔开

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

u = \frac{- ( - 10 ) + 2}{2 ( - 8 )} : - \frac{3}{4}

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

去除括号:

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

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

数字相加:

10 + 2 = 12

= \frac{12}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{12}{- 16}

使用分式法则:

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

= - \frac{12}{16}

约分:

4

= - \frac{3}{4}

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

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

去除括号:

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

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

数字相减:

10 - 2 = 8

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

数字相乘:

2 \cdot 8 = 16

= \frac{8}{- 16}

使用分式法则:

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

= - \frac{8}{16}

约分:

8

= - \frac{1}{2}

二次方程组的解是:

u = - \frac{3}{4}, u = - \frac{1}{2}

u = cos (x)

代回

cos (x) = - \frac{3}{4}, cos (x) = - \frac{1}{2}

cos (x) = - \frac{3}{4}, cos (x) = - \frac{1}{2}

cos (x) = - \frac{3}{4} : x = arccos (- \frac{3}{4}) + 2 πn, x = - arccos (- \frac{3}{4}) + 2 πn

cos (x) = - \frac{3}{4}

使用反三角函数性质

cos (x) = - \frac{3}{4}

cos (x) = - \frac{3}{4}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{3}{4}) + 2 πn, x = - arccos (- \frac{3}{4}) + 2 πn

x = arccos (- \frac{3}{4}) + 2 πn, x = - arccos (- \frac{3}{4}) + 2 πn

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = - \frac{1}{2}

cos (x) = - \frac{1}{2}

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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

合并所有解

x = arccos (- \frac{3}{4}) + 2 πn, x = - arccos (- \frac{3}{4}) + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

以小数形式表示解

x = 2.41885 \dots + 2 πn, x = - 2.41885 \dots + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

作图

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