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受欢迎的三角函数 >

25cos^2(θ)-9=0

解答

25 cos^{2} (θ) - 9 = 0

解答

θ = 0.92729 \dots + 2 πn, θ = 2 π - 0.92729 \dots + 2 πn, θ = 2.21429 \dots + 2 πn, θ = - 2.21429 \dots + 2 πn

+1

度数

θ = 53.13010 \dots^{\circ} + 36 0^{\circ} n, θ = 306.86989 \dots^{\circ} + 36 0^{\circ} n, θ = 126.86989 \dots^{\circ} + 36 0^{\circ} n, θ = - 126.86989 \dots^{\circ} + 36 0^{\circ} n

求解步骤

25 cos^{2} (θ) - 9 = 0

用替代法求解

25 cos^{2} (θ) - 9 = 0

令

: cos (θ) = u

25 u^{2} - 9 = 0

25 u^{2} - 9 = 0 : u = \frac{3}{5}, u = - \frac{3}{5}

25 u^{2} - 9 = 0

将

9

到右边

25 u^{2} - 9 = 0

两边加上

9

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

化简

25 u^{2} = 9

25 u^{2} = 9

两边除以

25

25 u^{2} = 9

两边除以

25

\frac{25 u ^{2}}{25} = \frac{9}{25}

化简

u^{2} = \frac{9}{25}

u^{2} = \frac{9}{25}

对于

x^{2} = f (a)

解为

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

u = \frac{9}{25}, u = - \frac{9}{25}

\frac{9}{25} = \frac{3}{5}

\frac{9}{25}

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{9}{25}

25 = 5

25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

= \frac{9}{5}

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{3}{5}

- \frac{9}{25} = - \frac{3}{5}

- \frac{9}{25}

化简

\frac{9}{25} : \frac{3}{5}

\frac{9}{25}

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{9}{25}

25 = 5

25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

= \frac{9}{5}

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{3}{5}

= - \frac{3}{5}

u = \frac{3}{5}, u = - \frac{3}{5}

u = cos (θ)

代回

cos (θ) = \frac{3}{5}, cos (θ) = - \frac{3}{5}

cos (θ) = \frac{3}{5}, cos (θ) = - \frac{3}{5}

cos (θ) = \frac{3}{5} : θ = arccos (\frac{3}{5}) + 2 πn, θ = 2 π - arccos (\frac{3}{5}) + 2 πn

cos (θ) = \frac{3}{5}

使用反三角函数性质

cos (θ) = \frac{3}{5}

cos (θ) = \frac{3}{5}

的通解

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

θ = arccos (\frac{3}{5}) + 2 πn, θ = 2 π - arccos (\frac{3}{5}) + 2 πn

θ = arccos (\frac{3}{5}) + 2 πn, θ = 2 π - arccos (\frac{3}{5}) + 2 πn

cos (θ) = - \frac{3}{5} : θ = arccos (- \frac{3}{5}) + 2 πn, θ = - arccos (- \frac{3}{5}) + 2 πn

cos (θ) = - \frac{3}{5}

使用反三角函数性质

cos (θ) = - \frac{3}{5}

cos (θ) = - \frac{3}{5}

的通解

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

θ = arccos (- \frac{3}{5}) + 2 πn, θ = - arccos (- \frac{3}{5}) + 2 πn

θ = arccos (- \frac{3}{5}) + 2 πn, θ = - arccos (- \frac{3}{5}) + 2 πn

合并所有解

θ = arccos (\frac{3}{5}) + 2 πn, θ = 2 π - arccos (\frac{3}{5}) + 2 πn, θ = arccos (- \frac{3}{5}) + 2 πn, θ = - arccos (- \frac{3}{5}) + 2 πn

以小数形式表示解

θ = 0.92729 \dots + 2 πn, θ = 2 π - 0.92729 \dots + 2 πn, θ = 2.21429 \dots + 2 πn, θ = - 2.21429 \dots + 2 πn

作图

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