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受欢迎的三角函数 >

1/9+cos^2(x)=1

解答

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

解答

x = 0.33983 \dots + 2 πn, x = 2 π - 0.33983 \dots + 2 πn, x = 2.80175 \dots + 2 πn, x = - 2.80175 \dots + 2 πn

+1

度数

x = 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 340.52877 \dots^{\circ} + 36 0^{\circ} n, x = 160.52877 \dots^{\circ} + 36 0^{\circ} n, x = - 160.52877 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: cos (x) = u

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

\frac{1}{9} + u^{2} = 1 : u = \frac{2 2}{3}, u = - \frac{2 2}{3}

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

将

\frac{1}{9}

到右边

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

两边减去

\frac{1}{9}

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

化简

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

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

化简

1 - \frac{1}{9} : \frac{8}{9}

1 - \frac{1}{9}

将项转换为分式:

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

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

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 9 - 1}{9}

1 \cdot 9 - 1 = 8

1 \cdot 9 - 1

数字相乘:

1 \cdot 9 = 9

= 9 - 1

数字相减:

9 - 1 = 8

= 8

= \frac{8}{9}

对于

x^{2} = f (a)

解为

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

u = \frac{8}{9}, u = - \frac{8}{9}

\frac{8}{9} = \frac{2 2}{3}

\frac{8}{9}

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{8}{9}

9 = 3

9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

= \frac{8}{3}

8 = 22

8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

= \frac{2 2}{3}

- \frac{8}{9} = - \frac{2 2}{3}

- \frac{8}{9}

化简

\frac{8}{9} : \frac{2 2}{3}

\frac{8}{9}

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{8}{9}

9 = 3

9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

= \frac{8}{3}

8 = 22

8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

= \frac{2 2}{3}

= - \frac{2 2}{3}

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

u = cos (x)

代回

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

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

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

cos (x) = \frac{2 2}{3}

使用反三角函数性质

cos (x) = \frac{2 2}{3}

cos (x) = \frac{2 2}{3}

的通解

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

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

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

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

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

使用反三角函数性质

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

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

的通解

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

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

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

合并所有解

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

以小数形式表示解

x = 0.33983 \dots + 2 πn, x = 2 π - 0.33983 \dots + 2 πn, x = 2.80175 \dots + 2 πn, x = - 2.80175 \dots + 2 πn

作图

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