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受欢迎的三角函数 >

8^{sin^2(x)}=4^{sin(x)-1/8}

解答

8^{s i n^{2} (x)} = 4^{s i n (x) - \frac{1}{8}}

解答

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = 0.16744 \dots + 2 πn, x = π - 0.16744 \dots + 2 πn

+1

度数

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 9.59406 \dots^{\circ} + 36 0^{\circ} n, x = 170.40593 \dots^{\circ} + 36 0^{\circ} n

求解步骤

8^{s i n^{2} (x)} = 4^{s i n (x) - \frac{1}{8}}

用替代法求解

8^{s i n^{2} (x)} = 4^{s i n (x) - \frac{1}{8}}

令

: sin (x) = u

8^{u^{2}} = 4^{u - \frac{1}{8}}

8^{u^{2}} = 4^{u - \frac{1}{8}} : u = \frac{1}{2}, u = \frac{1}{6}

8^{u^{2}} = 4^{u - \frac{1}{8}}

使用指数运算法则

8^{u^{2}} = 4^{u - \frac{1}{8}}

将转换为以

2

为底数的幂:

2^{3 u^{2}} = 2^{2 (u - \frac{1}{8})}

将

4

转换为以

2

为底数的幂

4 = 2^{2}

8^{u^{2}} = (2^{2})^{u - \frac{1}{8}}

将

8

转换为以

2

为底数的幂

8 = 2^{3}

(2^{3})^{u^{2}} = (2^{2})^{u - \frac{1}{8}}

使用指数法则:

(a^{b})^{c} = a^{b c}

(2^{3})^{u^{2}} = 2^{3 u^{2}}

2^{3 u^{2}} = (2^{2})^{u - \frac{1}{8}}

使用指数法则:

(a^{b})^{c} = a^{b c}

(2^{2})^{u - \frac{1}{8}} = 2^{2 (u - \frac{1}{8})}

2^{3 u^{2}} = 2^{2 (u - \frac{1}{8})}

2^{3 u^{2}} = 2^{2 (u - \frac{1}{8})}

若

a^{f (x)} = a^{g (x)}

，则

f (x) = g (x)

3 u^{2} = 2 (u - \frac{1}{8})

3 u^{2} = 2 (u - \frac{1}{8})

解

3 u^{2} = 2 (u - \frac{1}{8}) : u = \frac{1}{2}, u = \frac{1}{6}

3 u^{2} = 2 (u - \frac{1}{8})

展开

2 (u - \frac{1}{8}) : 2 u - \frac{1}{4}

2 (u - \frac{1}{8})

使用分配律:

a (b - c) = ab - a c

a = 2, b = u, c = \frac{1}{8}

= 2 u - 2 \cdot \frac{1}{8}

2 \cdot \frac{1}{8} = \frac{1}{4}

2 \cdot \frac{1}{8}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{8}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{8}

约分:

2

= \frac{1}{4}

= 2 u - \frac{1}{4}

3 u^{2} = 2 u - \frac{1}{4}

将

\frac{1}{4}

para o lado esquerdo

3 u^{2} = 2 u - \frac{1}{4}

两边加上

\frac{1}{4}

3 u^{2} + \frac{1}{4} = 2 u - \frac{1}{4} + \frac{1}{4}

化简

3 u^{2} + \frac{1}{4} = 2 u

3 u^{2} + \frac{1}{4} = 2 u

将

2 u

para o lado esquerdo

3 u^{2} + \frac{1}{4} = 2 u

两边减去

2 u

3 u^{2} + \frac{1}{4} - 2 u = 2 u - 2 u

化简

3 u^{2} + \frac{1}{4} - 2 u = 0

3 u^{2} + \frac{1}{4} - 2 u = 0

改写成标准形式

a x^{2} + b x + c = 0

3 u^{2} - 2 u + \frac{1}{4} = 0

使用求根公式求解

3 u^{2} - 2 u + \frac{1}{4} = 0

二次方程求根公式:

若

a = 3, b = - 2, c = \frac{1}{4}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 \cdot \frac{1}{4}}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 \cdot \frac{1}{4}}{2 \cdot 3}

(- 2)^{2} - 4 \cdot 3 \cdot \frac{1}{4} = 1

(- 2)^{2} - 4 \cdot 3 \cdot \frac{1}{4}

(- 2)^{2} = 2^{2}

(- 2)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 2)^{2} = 2^{2}

= 2^{2}

4 \cdot 3 \cdot \frac{1}{4} = 3

4 \cdot 3 \cdot \frac{1}{4}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 4 \cdot 3}{4}

约分:

4

= 1 \cdot 3

数字相乘:

1 \cdot 3 = 3

= 3

= 2^{2} - 3

2^{2} = 4

= 4 - 3

数字相减:

4 - 3 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 2 ) \pm 1}{2 \cdot 3}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 1}{2 \cdot 3}, u_{2} = \frac{- ( - 2 ) - 1}{2 \cdot 3}

u = \frac{- ( - 2 ) + 1}{2 \cdot 3} : \frac{1}{2}

\frac{- ( - 2 ) + 1}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{2 + 1}{2 \cdot 3}

数字相加:

2 + 1 = 3

= \frac{3}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{3}{6}

约分:

3

= \frac{1}{2}

u = \frac{- ( - 2 ) - 1}{2 \cdot 3} : \frac{1}{6}

\frac{- ( - 2 ) - 1}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{2 - 1}{2 \cdot 3}

数字相减:

2 - 1 = 1

= \frac{1}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{1}{6}

二次方程组的解是:

u = \frac{1}{2}, u = \frac{1}{6}

u = \frac{1}{2}, u = \frac{1}{6}

u = sin (x)

代回

sin (x) = \frac{1}{2}, sin (x) = \frac{1}{6}

sin (x) = \frac{1}{2}, sin (x) = \frac{1}{6}

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{6} : x = arcsin (\frac{1}{6}) + 2 πn, x = π - arcsin (\frac{1}{6}) + 2 πn

sin (x) = \frac{1}{6}

使用反三角函数性质

sin (x) = \frac{1}{6}

sin (x) = \frac{1}{6}

的通解

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

x = arcsin (\frac{1}{6}) + 2 πn, x = π - arcsin (\frac{1}{6}) + 2 πn

x = arcsin (\frac{1}{6}) + 2 πn, x = π - arcsin (\frac{1}{6}) + 2 πn

合并所有解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = arcsin (\frac{1}{6}) + 2 πn, x = π - arcsin (\frac{1}{6}) + 2 πn

以小数形式表示解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = 0.16744 \dots + 2 πn, x = π - 0.16744 \dots + 2 πn

作图

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