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受欢迎的三角函数 >

cos(2t)-cos(t)=-0.5,0<= t<= 2pi

解答

cos (2 t) - cos (t) = - 0.5, 0 \leq t \leq 2 π

解答

t = 0.62831 \dots, t = 2 π - 0.62831 \dots, t = 1.88495 \dots, t = - 1.88495 \dots + 2 π

+1

度数

t = 3 6^{\circ}, t = 32 4^{\circ}, t = 10 8^{\circ}, t = 25 2^{\circ}

求解步骤

cos (2 t) - cos (t) = - 0.5, 0 \leq t \leq 2 π

两边减去

- 0.5

cos (2 t) - cos (t) + 0.5 = 0

使用三角恒等式改写

0.5 + cos (2 t) - cos (t)

使用倍角公式:

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

= 0.5 + 2 cos^{2} (t) - 1 - cos (t)

化简

0.5 + 2 cos^{2} (t) - 1 - cos (t) : 2 cos^{2} (t) - cos (t) - 0.5

0.5 + 2 cos^{2} (t) - 1 - cos (t)

对同类项分组

= 2 cos^{2} (t) - cos (t) + 0.5 - 1

数字相加/相减:

0.5 - 1 = - 0.5

= 2 cos^{2} (t) - cos (t) - 0.5

= 2 cos^{2} (t) - cos (t) - 0.5

- 0.5 - cos (t) + 2 cos^{2} (t) = 0

用替代法求解

- 0.5 - cos (t) + 2 cos^{2} (t) = 0

令

: cos (t) = u

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

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

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

在两边乘以

10

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

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

- 0.5 \cdot 10 - u \cdot 10 + 2 u^{2} \cdot 10 = 0 \cdot 10

整理后得

- 5 - 10 u + 20 u^{2} = 0

- 5 - 10 u + 20 u^{2} = 0

改写成标准形式

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

20 u^{2} - 10 u - 5 = 0

使用求根公式求解

20 u^{2} - 10 u - 5 = 0

二次方程求根公式:

若

a = 20, b = - 10, c = - 5

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 20 ( - 5 )}{2 \cdot 20}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 20 ( - 5 )}{2 \cdot 20}

(- 10)^{2} - 4 \cdot 20 (- 5) = 105

(- 10)^{2} - 4 \cdot 20 (- 5)

使用法则

- (- a) = a

= (- 10)^{2} + 4 \cdot 20 \cdot 5

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 20 \cdot 5

数字相乘:

4 \cdot 20 \cdot 5 = 400

= 1 0^{2} + 400

1 0^{2} = 100

= 100 + 400

数字相加:

100 + 400 = 500

= 500

500

质因数分解:

2^{2} \cdot 5^{3}

500

500

除以

2500 = 250 \cdot 2

= 2 \cdot 250

250

除以

2250 = 125 \cdot 2

= 2 \cdot 2 \cdot 125

125

除以

5125 = 25 \cdot 5

= 2 \cdot 2 \cdot 5 \cdot 25

25

除以

525 = 5 \cdot 5

= 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5

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

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

使用指数法则:

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

= 2^{2} \cdot 5^{2} \cdot 5

使用根式运算法则:

n ab = n a n b

= 5 2^{2} 5^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 25 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 2 \cdot 55

整理后得

= 105

u_{1, 2} = \frac{- ( - 10 ) \pm 10 5}{2 \cdot 20}

将解分隔开

u_{1} = \frac{- ( - 10 ) + 10 5}{2 \cdot 20}, u_{2} = \frac{- ( - 10 ) - 10 5}{2 \cdot 20}

u = \frac{- ( - 10 ) + 10 5}{2 \cdot 20} : \frac{1 + 5}{4}

\frac{- ( - 10 ) + 10 5}{2 \cdot 20}

使用法则

- (- a) = a

= \frac{10 + 10 5}{2 \cdot 20}

数字相乘:

2 \cdot 20 = 40

= \frac{10 + 10 5}{40}

分解

10 + 105 : 10 (1 + 5)

10 + 105

改写为

= 10 \cdot 1 + 105

因式分解出通项

10

= 10 (1 + 5)

= \frac{10 ( 1 + 5 )}{40}

约分:

10

= \frac{1 + 5}{4}

u = \frac{- ( - 10 ) - 10 5}{2 \cdot 20} : \frac{1 - 5}{4}

\frac{- ( - 10 ) - 10 5}{2 \cdot 20}

使用法则

- (- a) = a

= \frac{10 - 10 5}{2 \cdot 20}

数字相乘:

2 \cdot 20 = 40

= \frac{10 - 10 5}{40}

分解

10 - 105 : 10 (1 - 5)

10 - 105

改写为

= 10 \cdot 1 - 105

因式分解出通项

10

= 10 (1 - 5)

= \frac{10 ( 1 - 5 )}{40}

约分:

10

= \frac{1 - 5}{4}

二次方程组的解是:

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

u = cos (t)

代回

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

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

cos (t) = \frac{1 + 5}{4}, 0 \leq t \leq 2 π : t = arccos (\frac{1 + 5}{4}), t = 2 π - arccos (\frac{1 + 5}{4})

cos (t) = \frac{1 + 5}{4}, 0 \leq t \leq 2 π

使用反三角函数性质

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

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

的通解

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

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

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

在

0 \leq t \leq 2 π

范围内的解

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

cos (t) = \frac{1 - 5}{4}, 0 \leq t \leq 2 π : t = arccos (\frac{1 - 5}{4}), t = - arccos (\frac{1 - 5}{4}) + 2 π

cos (t) = \frac{1 - 5}{4}, 0 \leq t \leq 2 π

使用反三角函数性质

cos (t) = \frac{1 - 5}{4}

cos (t) = \frac{1 - 5}{4}

的通解

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

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

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

在

0 \leq t \leq 2 π

范围内的解

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

合并所有解

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

以小数形式表示解

t = 0.62831 \dots, t = 2 π - 0.62831 \dots, t = 1.88495 \dots, t = - 1.88495 \dots + 2 π

作图

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