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受欢迎的三角函数 >

4sin^2(x)+2cos(x)+a=3

解答

4 sin^{2} (x) + 2 cos (x) + a = 3

解答

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

求解步骤

4 sin^{2} (x) + 2 cos (x) + a = 3

两边减去

3

4 sin^{2} (x) + 2 cos (x) + a - 3 = 0

使用三角恒等式改写

- 3 + a + 2 cos (x) + 4 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)

= - 3 + a + 2 cos (x) + 4 (1 - cos^{2} (x))

化简

- 3 + a + 2 cos (x) + 4 (1 - cos^{2} (x)) : 2 cos (x) - 4 cos^{2} (x) + a + 1

- 3 + a + 2 cos (x) + 4 (1 - cos^{2} (x))

乘开

4 (1 - cos^{2} (x)) : 4 - 4 cos^{2} (x)

4 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (x)

= 4 \cdot 1 - 4 cos^{2} (x)

数字相乘:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (x)

= - 3 + a + 2 cos (x) + 4 - 4 cos^{2} (x)

化简

- 3 + a + 2 cos (x) + 4 - 4 cos^{2} (x) : 2 cos (x) - 4 cos^{2} (x) + a + 1

- 3 + a + 2 cos (x) + 4 - 4 cos^{2} (x)

对同类项分组

= 2 cos (x) - 4 cos^{2} (x) + a - 3 + 4

数字相加/相减:

- 3 + 4 = 1

= 2 cos (x) - 4 cos^{2} (x) + a + 1

= 2 cos (x) - 4 cos^{2} (x) + a + 1

= 2 cos (x) - 4 cos^{2} (x) + a + 1

1 + a + 2 cos (x) - 4 cos^{2} (x) = 0

用替代法求解

1 + a + 2 cos (x) - 4 cos^{2} (x) = 0

令

: cos (x) = u

1 + a + 2 u - 4 u^{2} = 0

1 + a + 2 u - 4 u^{2} = 0 : u = - \frac{- 1 + 4 a + 5}{4}, u = \frac{4 a + 5 + 1}{4}

1 + a + 2 u - 4 u^{2} = 0

改写成标准形式

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

- 4 u^{2} + 2 u + 1 + a = 0

使用求根公式求解

- 4 u^{2} + 2 u + 1 + a = 0

二次方程求根公式:

若

a = - 4, b = 2, c = 1 + a

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 4 ) ( 1 + a )}{2 ( - 4 )}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 4 ) ( 1 + a )}{2 ( - 4 )}

化简

2^{2} - 4 (- 4) (1 + a) : 2 4 a + 5

2^{2} - 4 (- 4) (1 + a)

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 4 (1 + a)

数字相乘:

4 \cdot 4 = 16

= 2^{2} + 16 (a + 1)

分解

2^{2} + 16 (1 + a) : 4 (4 a + 5)

2^{2} + 16 (1 + a)

改写为

= 4 \cdot 1 + 4 \cdot 4 (1 + a)

因式分解出通项

4

= 4 (1 + 4 (1 + a))

乘开

4 (a + 1) + 1 : 4 a + 5

1 + 4 (1 + a)

乘开

4 (1 + a) : 4 + 4 a

4 (1 + a)

使用分配律:

a (b + c) = ab + a c

a = 4, b = 1, c = a

= 4 \cdot 1 + 4 a

数字相乘:

4 \cdot 1 = 4

= 4 + 4 a

= 1 + 4 + 4 a

数字相加:

1 + 4 = 5

= 4 a + 5

= 4 (4 a + 5)

= 4 (4 a + 5)

使用根式运算法则:

n ab = n a n b,

假定

a \geq 0, b \geq 0

= 4 4 a + 5

4 = 2

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

= 2 4 a + 5

u_{1, 2} = \frac{- 2 \pm 2 4 a + 5}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- 2 + 2 4 a + 5}{2 ( - 4 )}, u_{2} = \frac{- 2 - 2 4 a + 5}{2 ( - 4 )}

u = \frac{- 2 + 2 4 a + 5}{2 ( - 4 )} : - \frac{- 1 + 4 a + 5}{4}

\frac{- 2 + 2 4 a + 5}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 2 + 2 4 a + 5}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 2 + 2 4 a + 5}{- 8}

使用分式法则:

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

= - \frac{- 2 + 2 4 a + 5}{8}

消掉

\frac{- 2 + 2 4 a + 5}{8} : \frac{4 a + 5 - 1}{4}

\frac{- 2 + 2 4 a + 5}{8}

分解

- 2 + 2 4 a + 5 : 2 (- 1 + 5 + 4 a)

- 2 + 2 4 a + 5

改写为

= - 2 \cdot 1 + 2 5 + 4 a

因式分解出通项

2

= 2 (- 1 + 5 + 4 a)

= \frac{2 ( - 1 + 5 + 4 a )}{8}

约分:

2

= \frac{- 1 + 4 a + 5}{4}

= - \frac{4 a + 5 - 1}{4}

= - \frac{- 1 + 4 a + 5}{4}

u = \frac{- 2 - 2 4 a + 5}{2 ( - 4 )} : \frac{4 a + 5 + 1}{4}

\frac{- 2 - 2 4 a + 5}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 2 - 2 4 a + 5}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 2 - 2 4 a + 5}{- 8}

使用分式法则:

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

= - \frac{- 2 - 2 4 a + 5}{8}

消掉

\frac{- 2 - 2 4 a + 5}{8} : - \frac{4 a + 5 + 1}{4}

\frac{- 2 - 2 4 a + 5}{8}

分解

- 2 - 2 4 a + 5 : - 2 (1 + 5 + 4 a)

- 2 - 2 4 a + 5

改写为

= - 2 \cdot 1 - 2 5 + 4 a

因式分解出通项

2

= - 2 (1 + 5 + 4 a)

= - \frac{2 ( 1 + 5 + 4 a )}{8}

约分:

2

= - \frac{4 a + 5 + 1}{4}

= - (- \frac{4 a + 5 + 1}{4})

使用法则

- (- a) = a

= \frac{4 a + 5 + 1}{4}

二次方程组的解是:

u = - \frac{- 1 + 4 a + 5}{4}, u = \frac{4 a + 5 + 1}{4}

u = cos (x)

代回

cos (x) = - \frac{- 1 + 4 a + 5}{4}, cos (x) = \frac{4 a + 5 + 1}{4}

cos (x) = - \frac{- 1 + 4 a + 5}{4}, cos (x) = \frac{4 a + 5 + 1}{4}

cos (x) = - \frac{- 1 + 4 a + 5}{4} : x = arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn

cos (x) = - \frac{- 1 + 4 a + 5}{4}

使用反三角函数性质

cos (x) = - \frac{- 1 + 4 a + 5}{4}

cos (x) = - \frac{- 1 + 4 a + 5}{4}

的通解

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

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

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

cos (x) = \frac{4 a + 5 + 1}{4} : x = arccos (\frac{4 a + 5 + 1}{4}) + 2 πn, x = - arccos (\frac{4 a + 5 + 1}{4}) + 2 πn

cos (x) = \frac{4 a + 5 + 1}{4}

使用反三角函数性质

cos (x) = \frac{4 a + 5 + 1}{4}

cos (x) = \frac{4 a + 5 + 1}{4}

的通解

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

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

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

合并所有解

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

作图

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