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 >

cos^2(t)+9/49 =1

解答

cos^{2} (t) + \frac{9}{49} = 1

解答

t = 0.44291 \dots + 2 πn, t = 2 π - 0.44291 \dots + 2 πn, t = 2.69868 \dots + 2 πn, t = - 2.69868 \dots + 2 πn

+1

度数

t = 25.37693 \dots^{\circ} + 36 0^{\circ} n, t = 334.62306 \dots^{\circ} + 36 0^{\circ} n, t = 154.62306 \dots^{\circ} + 36 0^{\circ} n, t = - 154.62306 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cos^{2} (t) + \frac{9}{49} = 1

用替代法求解

cos^{2} (t) + \frac{9}{49} = 1

令

: cos (t) = u

u^{2} + \frac{9}{49} = 1

u^{2} + \frac{9}{49} = 1 : u = \frac{2 10}{7}, u = - \frac{2 10}{7}

u^{2} + \frac{9}{49} = 1

将

\frac{9}{49}

到右边

u^{2} + \frac{9}{49} = 1

两边减去

\frac{9}{49}

u^{2} + \frac{9}{49} - \frac{9}{49} = 1 - \frac{9}{49}

化简

u^{2} = 1 - \frac{9}{49}

u^{2} = 1 - \frac{9}{49}

化简

1 - \frac{9}{49} : \frac{40}{49}

1 - \frac{9}{49}

将项转换为分式:

1 = \frac{1 \cdot 49}{49}

= \frac{1 \cdot 49}{49} - \frac{9}{49}

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{1 \cdot 49 - 9}{49}

1 \cdot 49 - 9 = 40

1 \cdot 49 - 9

数字相乘:

1 \cdot 49 = 49

= 49 - 9

数字相减:

49 - 9 = 40

= 40

= \frac{40}{49}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = \frac{40}{49}, u = - \frac{40}{49}

\frac{40}{49} = \frac{2 10}{7}

\frac{40}{49}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{40}{49}

49 = 7

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

= \frac{40}{7}

40 = 210

40

40

质因数分解:

2^{3} \cdot 5

40

40

除以

240 = 20 \cdot 2

= 2 \cdot 20

20

除以

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 5

= 2^{3} \cdot 5

= 2^{3} \cdot 5

使用指数法则:

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

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

使用根式运算法则:

n ab = n a n b

= 2^{2} 2 \cdot 5

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 2 \cdot 5

整理后得

= 210

= \frac{2 10}{7}

- \frac{40}{49} = - \frac{2 10}{7}

- \frac{40}{49}

化简

\frac{40}{49} : \frac{2 10}{7}

\frac{40}{49}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{40}{49}

49 = 7

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

= \frac{40}{7}

40 = 210

40

40

质因数分解:

2^{3} \cdot 5

40

40

除以

240 = 20 \cdot 2

= 2 \cdot 20

20

除以

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 5

= 2^{3} \cdot 5

= 2^{3} \cdot 5

使用指数法则:

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

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

使用根式运算法则:

n ab = n a n b

= 2^{2} 2 \cdot 5

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 2 \cdot 5

整理后得

= 210

= \frac{2 10}{7}

= - \frac{2 10}{7}

u = \frac{2 10}{7}, u = - \frac{2 10}{7}

u = cos (t)

代回

cos (t) = \frac{2 10}{7}, cos (t) = - \frac{2 10}{7}

cos (t) = \frac{2 10}{7}, cos (t) = - \frac{2 10}{7}

cos (t) = \frac{2 10}{7} : t = arccos (\frac{2 10}{7}) + 2 πn, t = 2 π - arccos (\frac{2 10}{7}) + 2 πn

cos (t) = \frac{2 10}{7}

使用反三角函数性质

cos (t) = \frac{2 10}{7}

cos (t) = \frac{2 10}{7}

的通解

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

t = arccos (\frac{2 10}{7}) + 2 πn, t = 2 π - arccos (\frac{2 10}{7}) + 2 πn

t = arccos (\frac{2 10}{7}) + 2 πn, t = 2 π - arccos (\frac{2 10}{7}) + 2 πn

cos (t) = - \frac{2 10}{7} : t = arccos (- \frac{2 10}{7}) + 2 πn, t = - arccos (- \frac{2 10}{7}) + 2 πn

cos (t) = - \frac{2 10}{7}

使用反三角函数性质

cos (t) = - \frac{2 10}{7}

cos (t) = - \frac{2 10}{7}

的通解

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

t = arccos (- \frac{2 10}{7}) + 2 πn, t = - arccos (- \frac{2 10}{7}) + 2 πn

t = arccos (- \frac{2 10}{7}) + 2 πn, t = - arccos (- \frac{2 10}{7}) + 2 πn

合并所有解

t = arccos (\frac{2 10}{7}) + 2 πn, t = 2 π - arccos (\frac{2 10}{7}) + 2 πn, t = arccos (- \frac{2 10}{7}) + 2 πn, t = - arccos (- \frac{2 10}{7}) + 2 πn

以小数形式表示解

t = 0.44291 \dots + 2 πn, t = 2 π - 0.44291 \dots + 2 πn, t = 2.69868 \dots + 2 πn, t = - 2.69868 \dots + 2 πn

作图

流行的例子

4tan(x)+cot(x)=5

4 tan (x) + cot (x) = 5

证明 cos(pi/3+x)=sin(30-x)

p ro v e cos (\frac{π}{3} + x) = sin (3 0^{\circ} - x)

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

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

cos(θ)= 2/5 ,cot(θ)<0

cos (θ) = \frac{2}{5}, cot (θ) < 0

solvefor b,s(t)=tan(bt^2)

so l v e f or b, s (t) = tan (b t^{2})