zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

12sin^2(t)+cos(t)-1=5

解答

12 sin^{2} (t) + cos (t) - 1 = 5

解答

t = 2.30052 \dots + 2 πn, t = - 2.30052 \dots + 2 πn, t = 0.72273 \dots + 2 πn, t = 2 π - 0.72273 \dots + 2 πn

+1

度数

t = 131.81031 \dots^{\circ} + 36 0^{\circ} n, t = - 131.81031 \dots^{\circ} + 36 0^{\circ} n, t = 41.40962 \dots^{\circ} + 36 0^{\circ} n, t = 318.59037 \dots^{\circ} + 36 0^{\circ} n

求解步骤

12 sin^{2} (t) + cos (t) - 1 = 5

两边减去

5

12 sin^{2} (t) + cos (t) - 6 = 0

使用三角恒等式改写

- 6 + cos (t) + 12 sin^{2} (t)

使用毕达哥拉斯恒等式:

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

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

= - 6 + cos (t) + 12 (1 - cos^{2} (t))

化简

- 6 + cos (t) + 12 (1 - cos^{2} (t)) : cos (t) - 12 cos^{2} (t) + 6

- 6 + cos (t) + 12 (1 - cos^{2} (t))

乘开

12 (1 - cos^{2} (t)) : 12 - 12 cos^{2} (t)

12 (1 - cos^{2} (t))

使用分配律:

a (b - c) = ab - a c

a = 12, b = 1, c = cos^{2} (t)

= 12 \cdot 1 - 12 cos^{2} (t)

数字相乘:

12 \cdot 1 = 12

= 12 - 12 cos^{2} (t)

= - 6 + cos (t) + 12 - 12 cos^{2} (t)

化简

- 6 + cos (t) + 12 - 12 cos^{2} (t) : cos (t) - 12 cos^{2} (t) + 6

- 6 + cos (t) + 12 - 12 cos^{2} (t)

对同类项分组

= cos (t) - 12 cos^{2} (t) - 6 + 12

数字相加/相减:

- 6 + 12 = 6

= cos (t) - 12 cos^{2} (t) + 6

= cos (t) - 12 cos^{2} (t) + 6

= cos (t) - 12 cos^{2} (t) + 6

6 + cos (t) - 12 cos^{2} (t) = 0

用替代法求解

6 + cos (t) - 12 cos^{2} (t) = 0

令

: cos (t) = u

6 + u - 12 u^{2} = 0

6 + u - 12 u^{2} = 0 : u = - \frac{2}{3}, u = \frac{3}{4}

6 + u - 12 u^{2} = 0

改写成标准形式

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

- 12 u^{2} + u + 6 = 0

使用求根公式求解

- 12 u^{2} + u + 6 = 0

二次方程求根公式:

若

a = - 12, b = 1, c = 6

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 12 ) \cdot 6}{2 ( - 12 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 12 ) \cdot 6}{2 ( - 12 )}

1^{2} - 4 (- 12) \cdot 6 = 17

1^{2} - 4 (- 12) \cdot 6

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 12) \cdot 6

使用法则

- (- a) = a

= 1 + 4 \cdot 12 \cdot 6

数字相乘:

4 \cdot 12 \cdot 6 = 288

= 1 + 288

数字相加:

1 + 288 = 289

= 289

因式分解数字:

289 = 1 7^{2}

= 1 7^{2}

使用根式运算法则:

n a^{n} = a

1 7^{2} = 17

= 17

u_{1, 2} = \frac{- 1 \pm 17}{2 ( - 12 )}

将解分隔开

u_{1} = \frac{- 1 + 17}{2 ( - 12 )}, u_{2} = \frac{- 1 - 17}{2 ( - 12 )}

u = \frac{- 1 + 17}{2 ( - 12 )} : - \frac{2}{3}

\frac{- 1 + 17}{2 ( - 12 )}

去除括号:

(- a) = - a

= \frac{- 1 + 17}{- 2 \cdot 12}

数字相加/相减:

- 1 + 17 = 16

= \frac{16}{- 2 \cdot 12}

数字相乘:

2 \cdot 12 = 24

= \frac{16}{- 24}

使用分式法则:

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

= - \frac{16}{24}

约分:

8

= - \frac{2}{3}

u = \frac{- 1 - 17}{2 ( - 12 )} : \frac{3}{4}

\frac{- 1 - 17}{2 ( - 12 )}

去除括号:

(- a) = - a

= \frac{- 1 - 17}{- 2 \cdot 12}

数字相减:

- 1 - 17 = - 18

= \frac{- 18}{- 2 \cdot 12}

数字相乘:

2 \cdot 12 = 24

= \frac{- 18}{- 24}

使用分式法则:

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

= \frac{18}{24}

约分:

6

= \frac{3}{4}

二次方程组的解是:

u = - \frac{2}{3}, u = \frac{3}{4}

u = cos (t)

代回

cos (t) = - \frac{2}{3}, cos (t) = \frac{3}{4}

cos (t) = - \frac{2}{3}, cos (t) = \frac{3}{4}

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

cos (t) = - \frac{2}{3}

使用反三角函数性质

cos (t) = - \frac{2}{3}

cos (t) = - \frac{2}{3}

的通解

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

t = arccos (- \frac{2}{3}) + 2 πn, t = - arccos (- \frac{2}{3}) + 2 πn

t = arccos (- \frac{2}{3}) + 2 πn, t = - arccos (- \frac{2}{3}) + 2 πn

cos (t) = \frac{3}{4} : t = arccos (\frac{3}{4}) + 2 πn, t = 2 π - arccos (\frac{3}{4}) + 2 πn

cos (t) = \frac{3}{4}

使用反三角函数性质

cos (t) = \frac{3}{4}

cos (t) = \frac{3}{4}

的通解

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

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

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

合并所有解

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

以小数形式表示解

t = 2.30052 \dots + 2 πn, t = - 2.30052 \dots + 2 πn, t = 0.72273 \dots + 2 πn, t = 2 π - 0.72273 \dots + 2 πn

作图

流行的例子

sin (x) = \frac{5}{12}

arctan (x) = 2

tan(2x)= 1/(sqrt(3))

tan (2 x) = \frac{1}{3}

2sin(θ)cos(θ)-sqrt(3)cos(θ)=0

2 sin (θ) cos (θ) - 3 cos (θ) = 0

4sin(θ)cos(θ)+2sin(θ)-2cos(θ)-1=0

4 sin (θ) cos (θ) + 2 sin (θ) - 2 cos (θ) - 1 = 0