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受欢迎的三角函数 >

cos(x)(cos(x)+5)=2sin^2(x)

解答

cos (x) (cos (x) + 5) = 2 sin^{2} (x)

解答

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

+1

度数

x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cos (x) (cos (x) + 5) = 2 sin^{2} (x)

两边减去

2 sin^{2} (x)

cos (x) (cos (x) + 5) - 2 sin^{2} (x) = 0

使用三角恒等式改写

(5 + cos (x)) cos (x) - 2 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)

= (5 + cos (x)) cos (x) - 2 (1 - cos^{2} (x))

化简

(5 + cos (x)) cos (x) - 2 (1 - cos^{2} (x)) : 5 cos (x) + 3 cos^{2} (x) - 2

(5 + cos (x)) cos (x) - 2 (1 - cos^{2} (x))

= cos (x) (5 + cos (x)) - 2 (1 - cos^{2} (x))

乘开

cos (x) (5 + cos (x)) : 5 cos (x) + cos^{2} (x)

cos (x) (5 + cos (x))

使用分配律:

a (b + c) = ab + a c

a = cos (x), b = 5, c = cos (x)

= cos (x) \cdot 5 + cos (x) cos (x)

= 5 cos (x) + cos (x) cos (x)

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

cos (x) cos (x)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

= cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= cos^{2} (x)

= 5 cos (x) + cos^{2} (x)

= 5 cos (x) + cos^{2} (x) - 2 (1 - cos^{2} (x))

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

使用加减运算法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

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

= 5 cos (x) + cos^{2} (x) - 2 + 2 cos^{2} (x)

化简

5 cos (x) + cos^{2} (x) - 2 + 2 cos^{2} (x) : 5 cos (x) + 3 cos^{2} (x) - 2

5 cos (x) + cos^{2} (x) - 2 + 2 cos^{2} (x)

对同类项分组

= 5 cos (x) + cos^{2} (x) + 2 cos^{2} (x) - 2

同类项相加:

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

= 5 cos (x) + 3 cos^{2} (x) - 2

= 5 cos (x) + 3 cos^{2} (x) - 2

= 5 cos (x) + 3 cos^{2} (x) - 2

- 2 + 3 cos^{2} (x) + 5 cos (x) = 0

用替代法求解

- 2 + 3 cos^{2} (x) + 5 cos (x) = 0

令

: cos (x) = u

- 2 + 3 u^{2} + 5 u = 0

- 2 + 3 u^{2} + 5 u = 0 : u = \frac{1}{3}, u = - 2

- 2 + 3 u^{2} + 5 u = 0

改写成标准形式

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

3 u^{2} + 5 u - 2 = 0

使用求根公式求解

3 u^{2} + 5 u - 2 = 0

二次方程求根公式:

若

a = 3, b = 5, c = - 2

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 3 ( - 2 )}{2 \cdot 3}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 3 ( - 2 )}{2 \cdot 3}

5^{2} - 4 \cdot 3 (- 2) = 7

5^{2} - 4 \cdot 3 (- 2)

使用法则

- (- a) = a

= 5^{2} + 4 \cdot 3 \cdot 2

数字相乘:

4 \cdot 3 \cdot 2 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

数字相加:

25 + 24 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- 5 \pm 7}{2 \cdot 3}

将解分隔开

u_{1} = \frac{- 5 + 7}{2 \cdot 3}, u_{2} = \frac{- 5 - 7}{2 \cdot 3}

u = \frac{- 5 + 7}{2 \cdot 3} : \frac{1}{3}

\frac{- 5 + 7}{2 \cdot 3}

数字相加/相减:

- 5 + 7 = 2

= \frac{2}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{2}{6}

约分:

2

= \frac{1}{3}

u = \frac{- 5 - 7}{2 \cdot 3} : - 2

\frac{- 5 - 7}{2 \cdot 3}

数字相减:

- 5 - 7 = - 12

= \frac{- 12}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 12}{6}

使用分式法则:

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

= - \frac{12}{6}

数字相除:

\frac{12}{6} = 2

= - 2

二次方程组的解是:

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

u = cos (x)

代回

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

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

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

cos (x) = \frac{1}{3}

使用反三角函数性质

cos (x) = \frac{1}{3}

cos (x) = \frac{1}{3}

的通解

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

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

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

cos (x) = - 2 :

无解

cos (x) = - 2

- 1 \leq cos (x) \leq 1

无解

合并所有解

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

以小数形式表示解

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

作图

流行的例子

cos^2(x)-sin(x)=-1

cos^{2} (x) - sin (x) = - 1

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

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

cos^2(x)+3cos(x)=1

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

cot^2(x)=(tan(x))/2

cot^{2} (x) = \frac{tan ( x )}{2}

tan(θ)=(3.2)/(4.1)

tan (θ) = \frac{3.2}{4.1}