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受欢迎的三角函数 >

16tan^2(θ)-1=0

解答

16 tan^{2} (θ) - 1 = 0

解答

θ = 0.24497 \dots + πn, θ = - 0.24497 \dots + πn

+1

度数

θ = 14.03624 \dots^{\circ} + 18 0^{\circ} n, θ = - 14.03624 \dots^{\circ} + 18 0^{\circ} n

求解步骤

16 tan^{2} (θ) - 1 = 0

用替代法求解

16 tan^{2} (θ) - 1 = 0

令

: tan (θ) = u

16 u^{2} - 1 = 0

16 u^{2} - 1 = 0 : u = \frac{1}{4}, u = - \frac{1}{4}

16 u^{2} - 1 = 0

将

1

到右边

16 u^{2} - 1 = 0

两边加上

1

16 u^{2} - 1 + 1 = 0 + 1

化简

16 u^{2} = 1

16 u^{2} = 1

两边除以

16

16 u^{2} = 1

两边除以

16

\frac{16 u ^{2}}{16} = \frac{1}{16}

化简

u^{2} = \frac{1}{16}

u^{2} = \frac{1}{16}

对于

x^{2} = f (a)

解为

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

u = \frac{1}{16}, u = - \frac{1}{16}

\frac{1}{16} = \frac{1}{4}

\frac{1}{16}

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}{16}

16 = 4

16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a

4^{2} = 4

= 4

= \frac{1}{4}

使用法则

1 = 1

= \frac{1}{4}

- \frac{1}{16} = - \frac{1}{4}

- \frac{1}{16}

化简

\frac{1}{16} : \frac{1}{4}

\frac{1}{16}

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}{16}

16 = 4

16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a

4^{2} = 4

= 4

= \frac{1}{4}

= - \frac{1}{4}

使用法则

1 = 1

= - \frac{1}{4}

u = \frac{1}{4}, u = - \frac{1}{4}

u = tan (θ)

代回

tan (θ) = \frac{1}{4}, tan (θ) = - \frac{1}{4}

tan (θ) = \frac{1}{4}, tan (θ) = - \frac{1}{4}

tan (θ) = \frac{1}{4} : θ = arctan (\frac{1}{4}) + πn

tan (θ) = \frac{1}{4}

使用反三角函数性质

tan (θ) = \frac{1}{4}

tan (θ) = \frac{1}{4}

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

θ = arctan (\frac{1}{4}) + πn

θ = arctan (\frac{1}{4}) + πn

tan (θ) = - \frac{1}{4} : θ = arctan (- \frac{1}{4}) + πn

tan (θ) = - \frac{1}{4}

使用反三角函数性质

tan (θ) = - \frac{1}{4}

tan (θ) = - \frac{1}{4}

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

θ = arctan (- \frac{1}{4}) + πn

θ = arctan (- \frac{1}{4}) + πn

合并所有解

θ = arctan (\frac{1}{4}) + πn, θ = arctan (- \frac{1}{4}) + πn

以小数形式表示解

θ = 0.24497 \dots + πn, θ = - 0.24497 \dots + πn

作图

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