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受欢迎的三角函数 >

36/49+cos^2(θ)=1

解答

\frac{36}{49} + cos^{2} (θ) = 1

解答

θ = 1.02969 \dots + 2 πn, θ = 2 π - 1.02969 \dots + 2 πn, θ = 2.11189 \dots + 2 πn, θ = - 2.11189 \dots + 2 πn

+1

度数

θ = 58.99728 \dots^{\circ} + 36 0^{\circ} n, θ = 301.00271 \dots^{\circ} + 36 0^{\circ} n, θ = 121.00271 \dots^{\circ} + 36 0^{\circ} n, θ = - 121.00271 \dots^{\circ} + 36 0^{\circ} n

求解步骤

\frac{36}{49} + cos^{2} (θ) = 1

用替代法求解

\frac{36}{49} + cos^{2} (θ) = 1

令

: cos (θ) = u

\frac{36}{49} + u^{2} = 1

\frac{36}{49} + u^{2} = 1 : u = \frac{13}{7}, u = - \frac{13}{7}

\frac{36}{49} + u^{2} = 1

将

\frac{36}{49}

到右边

\frac{36}{49} + u^{2} = 1

两边减去

\frac{36}{49}

\frac{36}{49} + u^{2} - \frac{36}{49} = 1 - \frac{36}{49}

化简

u^{2} = 1 - \frac{36}{49}

u^{2} = 1 - \frac{36}{49}

化简

1 - \frac{36}{49} : \frac{13}{49}

1 - \frac{36}{49}

将项转换为分式:

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

= \frac{1 \cdot 49}{49} - \frac{36}{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 - 36}{49}

1 \cdot 49 - 36 = 13

1 \cdot 49 - 36

数字相乘:

1 \cdot 49 = 49

= 49 - 36

数字相减:

49 - 36 = 13

= 13

= \frac{13}{49}

对于

x^{2} = f (a)

解为

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

u = \frac{13}{49}, u = - \frac{13}{49}

\frac{13}{49} = \frac{13}{7}

\frac{13}{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{13}{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{13}{7}

- \frac{13}{49} = - \frac{13}{7}

- \frac{13}{49}

化简

\frac{13}{49} : \frac{13}{7}

\frac{13}{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{13}{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{13}{7}

= - \frac{13}{7}

u = \frac{13}{7}, u = - \frac{13}{7}

u = cos (θ)

代回

cos (θ) = \frac{13}{7}, cos (θ) = - \frac{13}{7}

cos (θ) = \frac{13}{7}, cos (θ) = - \frac{13}{7}

cos (θ) = \frac{13}{7} : θ = arccos (\frac{13}{7}) + 2 πn, θ = 2 π - arccos (\frac{13}{7}) + 2 πn

cos (θ) = \frac{13}{7}

使用反三角函数性质

cos (θ) = \frac{13}{7}

cos (θ) = \frac{13}{7}

的通解

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

θ = arccos (\frac{13}{7}) + 2 πn, θ = 2 π - arccos (\frac{13}{7}) + 2 πn

θ = arccos (\frac{13}{7}) + 2 πn, θ = 2 π - arccos (\frac{13}{7}) + 2 πn

cos (θ) = - \frac{13}{7} : θ = arccos (- \frac{13}{7}) + 2 πn, θ = - arccos (- \frac{13}{7}) + 2 πn

cos (θ) = - \frac{13}{7}

使用反三角函数性质

cos (θ) = - \frac{13}{7}

cos (θ) = - \frac{13}{7}

的通解

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

θ = arccos (- \frac{13}{7}) + 2 πn, θ = - arccos (- \frac{13}{7}) + 2 πn

θ = arccos (- \frac{13}{7}) + 2 πn, θ = - arccos (- \frac{13}{7}) + 2 πn

合并所有解

θ = arccos (\frac{13}{7}) + 2 πn, θ = 2 π - arccos (\frac{13}{7}) + 2 πn, θ = arccos (- \frac{13}{7}) + 2 πn, θ = - arccos (- \frac{13}{7}) + 2 πn

以小数形式表示解

θ = 1.02969 \dots + 2 πn, θ = 2 π - 1.02969 \dots + 2 πn, θ = 2.11189 \dots + 2 πn, θ = - 2.11189 \dots + 2 πn

作图

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