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受欢迎的三角函数 >

2+cos(x)=3(cos^2(x))/2+(sin^2(x))/2

解答

2 + cos (x) = 3 \frac{cos ^{2} ( x )}{2} + \frac{sin ^{2} ( x )}{2}

解答

x = 2.53724 \dots + 2 πn, x = - 2.53724 \dots + 2 πn

+1

度数

x = 145.37370 \dots^{\circ} + 36 0^{\circ} n, x = - 145.37370 \dots^{\circ} + 36 0^{\circ} n

求解步骤

2 + cos (x) = 3 \cdot \frac{cos ^{2} ( x )}{2} + \frac{sin ^{2} ( x )}{2}

两边减去

3 \frac{cos ^{2} ( x )}{2} + \frac{sin ^{2} ( x )}{2}

2 + cos (x) - \frac{3 cos ^{2} ( x ) + sin ^{2} ( x )}{2} = 0

化简

2 + cos (x) - \frac{3 cos ^{2} ( x ) + sin ^{2} ( x )}{2} : \frac{4 + 2 cos ( x ) - 3 cos ^{2} ( x ) - sin ^{2} ( x )}{2}

2 + cos (x) - \frac{3 cos ^{2} ( x ) + sin ^{2} ( x )}{2}

将项转换为分式:

2 = \frac{2 \cdot 2}{2}, cos (x) = \frac{cos ( x ) 2}{2}

= \frac{2 \cdot 2}{2} + \frac{cos ( x ) \cdot 2}{2} - \frac{3 cos ^{2} ( x ) + sin ^{2} ( x )}{2}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 \cdot 2 + cos ( x ) \cdot 2 - ( 3 cos ^{2} ( x ) + sin ^{2} ( x ) )}{2}

数字相乘:

2 \cdot 2 = 4

= \frac{4 + 2 cos ( x ) - ( 3 cos ^{2} ( x ) + sin ^{2} ( x ) )}{2}

乘开

4 + cos (x) \cdot 2 - (3 cos^{2} (x) + sin^{2} (x)) : 4 + cos (x) \cdot 2 - 3 cos^{2} (x) - sin^{2} (x)

4 + cos (x) \cdot 2 - (3 cos^{2} (x) + sin^{2} (x))

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

- (3 cos^{2} (x) + sin^{2} (x)) : - 3 cos^{2} (x) - sin^{2} (x)

- (3 cos^{2} (x) + sin^{2} (x))

打开括号

= - (3 cos^{2} (x)) - (sin^{2} (x))

使用加减运算法则

+ (- a) = - a

= - 3 cos^{2} (x) - sin^{2} (x)

= 4 + cos (x) \cdot 2 - 3 cos^{2} (x) - sin^{2} (x)

= \frac{4 + 2 cos ( x ) - 3 cos ^{2} ( x ) - sin ^{2} ( x )}{2}

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

使用三角恒等式改写

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

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

化简

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

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

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

化简

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

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

对同类项分组

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

同类项相加:

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

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

数字相加/相减:

4 - 1 = 3

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

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

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

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

用替代法求解

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

令

: cos (x) = u

3 + 2 u - 2 u^{2} = 0

3 + 2 u - 2 u^{2} = 0 : u = - \frac{- 1 + 7}{2}, u = \frac{1 + 7}{2}

3 + 2 u - 2 u^{2} = 0

改写成标准形式

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

- 2 u^{2} + 2 u + 3 = 0

使用求根公式求解

- 2 u^{2} + 2 u + 3 = 0

二次方程求根公式:

若

a = - 2, b = 2, c = 3

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

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

2^{2} - 4 (- 2) \cdot 3 = 27

2^{2} - 4 (- 2) \cdot 3

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 2 \cdot 3

数字相乘:

4 \cdot 2 \cdot 3 = 24

= 2^{2} + 24

2^{2} = 4

= 4 + 24

数字相加:

4 + 24 = 28

= 28

28

质因数分解:

2^{2} \cdot 7

28

28

除以

228 = 14 \cdot 2

= 2 \cdot 14

14

除以

214 = 7 \cdot 2

= 2 \cdot 2 \cdot 7

2, 7

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 7

= 2^{2} \cdot 7

= 2^{2} \cdot 7

使用根式运算法则:

n ab = n a n b

= 7 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 27

u_{1, 2} = \frac{- 2 \pm 2 7}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 2 + 2 7}{2 ( - 2 )}, u_{2} = \frac{- 2 - 2 7}{2 ( - 2 )}

u = \frac{- 2 + 2 7}{2 ( - 2 )} : - \frac{- 1 + 7}{2}

\frac{- 2 + 2 7}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 2 + 2 7}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 + 2 7}{- 4}

使用分式法则:

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

= - \frac{- 2 + 2 7}{4}

消掉

\frac{- 2 + 2 7}{4} : \frac{7 - 1}{2}

\frac{- 2 + 2 7}{4}

分解

- 2 + 27 : 2 (- 1 + 7)

- 2 + 27

改写为

= - 2 \cdot 1 + 27

因式分解出通项

2

= 2 (- 1 + 7)

= \frac{2 ( - 1 + 7 )}{4}

约分:

2

= \frac{- 1 + 7}{2}

= - \frac{7 - 1}{2}

= - \frac{- 1 + 7}{2}

u = \frac{- 2 - 2 7}{2 ( - 2 )} : \frac{1 + 7}{2}

\frac{- 2 - 2 7}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 2 - 2 7}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 - 2 7}{- 4}

使用分式法则:

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

- 2 - 27 = - (2 + 27)

= \frac{2 + 2 7}{4}

分解

2 + 27 : 2 (1 + 7)

2 + 27

改写为

= 2 \cdot 1 + 27

因式分解出通项

2

= 2 (1 + 7)

= \frac{2 ( 1 + 7 )}{4}

约分:

2

= \frac{1 + 7}{2}

二次方程组的解是:

u = - \frac{- 1 + 7}{2}, u = \frac{1 + 7}{2}

u = cos (x)

代回

cos (x) = - \frac{- 1 + 7}{2}, cos (x) = \frac{1 + 7}{2}

cos (x) = - \frac{- 1 + 7}{2}, cos (x) = \frac{1 + 7}{2}

cos (x) = - \frac{- 1 + 7}{2} : x = arccos (- \frac{- 1 + 7}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 7}{2}) + 2 πn

cos (x) = - \frac{- 1 + 7}{2}

使用反三角函数性质

cos (x) = - \frac{- 1 + 7}{2}

cos (x) = - \frac{- 1 + 7}{2}

的通解

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

x = arccos (- \frac{- 1 + 7}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 7}{2}) + 2 πn

x = arccos (- \frac{- 1 + 7}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 7}{2}) + 2 πn

cos (x) = \frac{1 + 7}{2} :

无解

cos (x) = \frac{1 + 7}{2}

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = arccos (- \frac{- 1 + 7}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 7}{2}) + 2 πn

以小数形式表示解

x = 2.53724 \dots + 2 πn, x = - 2.53724 \dots + 2 πn

作图

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