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受欢迎的三角函数 >

(tan(θ)-2)(16sin^2(θ)-1)=0

解答

(tan (θ) - 2) (16 sin^{2} (θ) - 1) = 0

解答

θ = 1.10714 \dots + πn, θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn, θ = - 0.25268 \dots + 2 πn, θ = π + 0.25268 \dots + 2 πn

+1

度数

θ = 63.43494 \dots^{\circ} + 18 0^{\circ} n, θ = 14.47751 \dots^{\circ} + 36 0^{\circ} n, θ = 165.52248 \dots^{\circ} + 36 0^{\circ} n, θ = - 14.47751 \dots^{\circ} + 36 0^{\circ} n, θ = 194.47751 \dots^{\circ} + 36 0^{\circ} n

求解步骤

(tan (θ) - 2) (16 sin^{2} (θ) - 1) = 0

分别求解每个部分

tan (θ) - 2 = 0 or 16 sin^{2} (θ) - 1 = 0

tan (θ) - 2 = 0 : θ = arctan (2) + πn

tan (θ) - 2 = 0

将

2

到右边

tan (θ) - 2 = 0

两边加上

2

tan (θ) - 2 + 2 = 0 + 2

化简

tan (θ) = 2

tan (θ) = 2

使用反三角函数性质

tan (θ) = 2

tan (θ) = 2

的通解

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

θ = arctan (2) + πn

θ = arctan (2) + πn

16 sin^{2} (θ) - 1 = 0 : θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn, θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

16 sin^{2} (θ) - 1 = 0

用替代法求解

16 sin^{2} (θ) - 1 = 0

令

: sin (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 = sin (θ)

代回

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

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

sin (θ) = \frac{1}{4} : θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

sin (θ) = \frac{1}{4}

使用反三角函数性质

sin (θ) = \frac{1}{4}

sin (θ) = \frac{1}{4}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

sin (θ) = - \frac{1}{4} : θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

sin (θ) = - \frac{1}{4}

使用反三角函数性质

sin (θ) = - \frac{1}{4}

sin (θ) = - \frac{1}{4}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

合并所有解

θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn, θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

合并所有解

θ = arctan (2) + πn, θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn, θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

以小数形式表示解

θ = 1.10714 \dots + πn, θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn, θ = - 0.25268 \dots + 2 πn, θ = π + 0.25268 \dots + 2 πn

作图

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