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受欢迎的三角函数 >

9cos^2(θ)-4=0

解答

9 cos^{2} (θ) - 4 = 0

解答

θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn, θ = 2.30052 \dots + 2 πn, θ = - 2.30052 \dots + 2 πn

+1

度数

θ = 48.18968 \dots^{\circ} + 36 0^{\circ} n, θ = 311.81031 \dots^{\circ} + 36 0^{\circ} n, θ = 131.81031 \dots^{\circ} + 36 0^{\circ} n, θ = - 131.81031 \dots^{\circ} + 36 0^{\circ} n

求解步骤

9 cos^{2} (θ) - 4 = 0

用替代法求解

9 cos^{2} (θ) - 4 = 0

令

: cos (θ) = u

9 u^{2} - 4 = 0

9 u^{2} - 4 = 0 : u = \frac{2}{3}, u = - \frac{2}{3}

9 u^{2} - 4 = 0

将

4

到右边

9 u^{2} - 4 = 0

两边加上

4

9 u^{2} - 4 + 4 = 0 + 4

化简

9 u^{2} = 4

9 u^{2} = 4

两边除以

9

9 u^{2} = 4

两边除以

9

\frac{9 u ^{2}}{9} = \frac{4}{9}

化简

u^{2} = \frac{4}{9}

u^{2} = \frac{4}{9}

对于

x^{2} = f (a)

解为

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

u = \frac{4}{9}, u = - \frac{4}{9}

\frac{4}{9} = \frac{2}{3}

\frac{4}{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{4}{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{4}{3}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2}{3}

- \frac{4}{9} = - \frac{2}{3}

- \frac{4}{9}

化简

\frac{4}{9} : \frac{2}{3}

\frac{4}{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{4}{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{4}{3}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2}{3}

= - \frac{2}{3}

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

u = cos (θ)

代回

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

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

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

cos (θ) = \frac{2}{3}

使用反三角函数性质

cos (θ) = \frac{2}{3}

cos (θ) = \frac{2}{3}

的通解

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

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

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

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

cos (θ) = - \frac{2}{3}

使用反三角函数性质

cos (θ) = - \frac{2}{3}

cos (θ) = - \frac{2}{3}

的通解

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

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

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

合并所有解

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

以小数形式表示解

θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn, θ = 2.30052 \dots + 2 πn, θ = - 2.30052 \dots + 2 πn

作图

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