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受欢迎的三角函数 >

8sin^2(x)+2cos(x)=7

解答

8 sin^{2} (x) + 2 cos (x) = 7

解答

x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

+1

度数

x = 104.47751 \dots^{\circ} + 36 0^{\circ} n, x = - 104.47751 \dots^{\circ} + 36 0^{\circ} n, x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

求解步骤

8 sin^{2} (x) + 2 cos (x) = 7

两边减去

7

8 sin^{2} (x) + 2 cos (x) - 7 = 0

使用三角恒等式改写

- 7 + 2 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)

= - 7 + 2 cos (x) + 8 (1 - cos^{2} (x))

化简

- 7 + 2 cos (x) + 8 (1 - cos^{2} (x)) : 2 cos (x) - 8 cos^{2} (x) + 1

- 7 + 2 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)

= - 7 + 2 cos (x) + 8 - 8 cos^{2} (x)

化简

- 7 + 2 cos (x) + 8 - 8 cos^{2} (x) : 2 cos (x) - 8 cos^{2} (x) + 1

- 7 + 2 cos (x) + 8 - 8 cos^{2} (x)

对同类项分组

= 2 cos (x) - 8 cos^{2} (x) - 7 + 8

数字相加/相减:

- 7 + 8 = 1

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

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

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

1 + 2 cos (x) - 8 cos^{2} (x) = 0

用替代法求解

1 + 2 cos (x) - 8 cos^{2} (x) = 0

令

: cos (x) = u

1 + 2 u - 8 u^{2} = 0

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

1 + 2 u - 8 u^{2} = 0

改写成标准形式

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

- 8 u^{2} + 2 u + 1 = 0

使用求根公式求解

- 8 u^{2} + 2 u + 1 = 0

二次方程求根公式:

若

a = - 8, b = 2, c = 1

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

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

2^{2} - 4 (- 8) \cdot 1 = 6

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

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 8 \cdot 1

数字相乘:

4 \cdot 8 \cdot 1 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

数字相加:

4 + 32 = 36

= 36

因式分解数字:

36 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- 2 \pm 6}{2 ( - 8 )}

将解分隔开

u_{1} = \frac{- 2 + 6}{2 ( - 8 )}, u_{2} = \frac{- 2 - 6}{2 ( - 8 )}

u = \frac{- 2 + 6}{2 ( - 8 )} : - \frac{1}{4}

\frac{- 2 + 6}{2 ( - 8 )}

去除括号:

(- a) = - a

= \frac{- 2 + 6}{- 2 \cdot 8}

数字相加/相减:

- 2 + 6 = 4

= \frac{4}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{4}{- 16}

使用分式法则:

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

= - \frac{4}{16}

约分:

4

= - \frac{1}{4}

u = \frac{- 2 - 6}{2 ( - 8 )} : \frac{1}{2}

\frac{- 2 - 6}{2 ( - 8 )}

去除括号:

(- a) = - a

= \frac{- 2 - 6}{- 2 \cdot 8}

数字相减:

- 2 - 6 = - 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{1}{4}, u = \frac{1}{2}

u = cos (x)

代回

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

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

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

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

使用反三角函数性质

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

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

的通解

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

x = arccos (- \frac{1}{4}) + 2 πn, x = - arccos (- \frac{1}{4}) + 2 πn

x = arccos (- \frac{1}{4}) + 2 πn, x = - arccos (- \frac{1}{4}) + 2 πn

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 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{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

合并所有解

x = arccos (- \frac{1}{4}) + 2 πn, x = - arccos (- \frac{1}{4}) + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

以小数形式表示解

x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

作图

流行的例子

tan((19pi)/(12))

tan (\frac{19 π}{12})

证明 1/(sec(x)-tan(x))=sec(x)+tan(x)

p ro v e \frac{1}{sec ( x ) - tan ( x )} = sec (x) + tan (x)

cos (9 0^{\circ})

cos (- \frac{3 π}{4})

sin (- \frac{5 π}{6})