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受欢迎的三角函数 >

16cos^2(θ)-2=0

解答

16 cos^{2} (θ) - 2 = 0

解答

θ = 1.20942 \dots + 2 πn, θ = 2 π - 1.20942 \dots + 2 πn, θ = 1.93216 \dots + 2 πn, θ = - 1.93216 \dots + 2 πn

+1

度数

θ = 69.29518 \dots^{\circ} + 36 0^{\circ} n, θ = 290.70481 \dots^{\circ} + 36 0^{\circ} n, θ = 110.70481 \dots^{\circ} + 36 0^{\circ} n, θ = - 110.70481 \dots^{\circ} + 36 0^{\circ} n

求解步骤

16 cos^{2} (θ) - 2 = 0

用替代法求解

16 cos^{2} (θ) - 2 = 0

令

: cos (θ) = u

16 u^{2} - 2 = 0

16 u^{2} - 2 = 0 : u = \frac{2}{4}, u = - \frac{2}{4}

16 u^{2} - 2 = 0

将

2

到右边

16 u^{2} - 2 = 0

两边加上

2

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

化简

16 u^{2} = 2

16 u^{2} = 2

两边除以

16

16 u^{2} = 2

两边除以

16

\frac{16 u ^{2}}{16} = \frac{2}{16}

化简

u^{2} = \frac{1}{8}

u^{2} = \frac{1}{8}

对于

x^{2} = f (a)

解为

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

u = \frac{1}{8}, u = - \frac{1}{8}

\frac{1}{8} = \frac{2}{4}

\frac{1}{8}

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{1}{8}

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{1}{2 2}

使用法则

1 = 1

= \frac{1}{2 2}

\frac{1}{2 2}

有理化:

\frac{2}{4}

\frac{1}{2 2}

乘以共轭根式

\frac{2}{2}

= \frac{1 \cdot 2}{2 2 2}

1 \cdot 2 = 2

222 = 4

222

使用指数法则:

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

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

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

同类项相加:

\frac{1}{2} + \frac{1}{2} = 2 \cdot \frac{1}{2}

= 2^{1 + 2 \cdot \frac{1}{2}}

2 \cdot \frac{1}{2} = 1

2 \cdot \frac{1}{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2}{4}

= \frac{2}{4}

- \frac{1}{8} = - \frac{2}{4}

- \frac{1}{8}

化简

\frac{1}{8} : \frac{1}{2 2}

\frac{1}{8}

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{1}{8}

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{1}{2 2}

= - \frac{1}{2 2}

使用法则

1 = 1

= - \frac{1}{2 2}

- \frac{1}{2 2}

有理化:

- \frac{2}{4}

- \frac{1}{2 2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

222 = 4

222

使用指数法则:

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

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

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

同类项相加:

\frac{1}{2} + \frac{1}{2} = 2 \cdot \frac{1}{2}

= 2^{1 + 2 \cdot \frac{1}{2}}

2 \cdot \frac{1}{2} = 1

2 \cdot \frac{1}{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= - \frac{2}{4}

= - \frac{2}{4}

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

u = cos (θ)

代回

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

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

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

cos (θ) = \frac{2}{4}

使用反三角函数性质

cos (θ) = \frac{2}{4}

cos (θ) = \frac{2}{4}

的通解

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

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

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

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

cos (θ) = - \frac{2}{4}

使用反三角函数性质

cos (θ) = - \frac{2}{4}

cos (θ) = - \frac{2}{4}

的通解

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

θ = arccos (- \frac{2}{4}) + 2 πn, θ = - arccos (- \frac{2}{4}) + 2 πn

θ = arccos (- \frac{2}{4}) + 2 πn, θ = - arccos (- \frac{2}{4}) + 2 πn

合并所有解

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

以小数形式表示解

θ = 1.20942 \dots + 2 πn, θ = 2 π - 1.20942 \dots + 2 πn, θ = 1.93216 \dots + 2 πn, θ = - 1.93216 \dots + 2 πn

作图

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