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受欢迎的三角函数 >

cos^2(x)+sin^2(2x)=0

解答

cos^{2} (x) + sin^{2} (2 x) = 0

解答

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

cos^{2} (x) + sin^{2} (2 x) = 0

使用三角恒等式改写

cos^{2} (x) + sin^{2} (2 x)

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

= cos^{2} (x) + (2 sin (x) cos (x))^{2}

化简

cos^{2} (x) + (2 sin (x) cos (x))^{2} : cos^{2} (x) + 4 sin^{2} (x) cos^{2} (x)

cos^{2} (x) + (2 sin (x) cos (x))^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= cos^{2} (x) + 2^{2} sin^{2} (x) cos^{2} (x)

2^{2} = 4

= cos^{2} (x) + 4 sin^{2} (x) cos^{2} (x)

= cos^{2} (x) + 4 sin^{2} (x) cos^{2} (x)

cos^{2} (x) + 4 cos^{2} (x) sin^{2} (x) = 0

分解

cos^{2} (x) + 4 cos^{2} (x) sin^{2} (x) : cos^{2} (x) (4 sin^{2} (x) + 1)

cos^{2} (x) + 4 cos^{2} (x) sin^{2} (x)

因式分解出通项

cos^{2} (x)

= cos^{2} (x) (1 + 4 sin^{2} (x))

cos^{2} (x) (4 sin^{2} (x) + 1) = 0

分别求解每个部分

cos^{2} (x) = 0 or 4 sin^{2} (x) + 1 = 0

cos^{2} (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos^{2} (x) = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

cos (x) = 0

cos (x) = 0

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

4 sin^{2} (x) + 1 = 0 :

无解

4 sin^{2} (x) + 1 = 0

用替代法求解

4 sin^{2} (x) + 1 = 0

令

: sin (x) = u

4 u^{2} + 1 = 0

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

4 u^{2} + 1 = 0

将

1

到右边

4 u^{2} + 1 = 0

两边减去

1

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

化简

4 u^{2} = - 1

4 u^{2} = - 1

两边除以

4

4 u^{2} = - 1

两边除以

4

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

化简

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

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

对于

x^{2} = f (a)

解为

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

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

化简

- \frac{1}{4} : i \frac{1}{2}

- \frac{1}{4}

使用根式运算法则:

- a = - 1 a

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

= - 1 \frac{1}{4}

使用虚数运算法则:

- 1 = i

= i \frac{1}{4}

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

= i \frac{1}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= i \frac{1}{2}

使用法则

1 = 1

= i \frac{1}{2}

将

i \frac{1}{2}

改写成标准复数形式:

\frac{1}{2} i

i \frac{1}{2}

分式相乘:

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

= \frac{1 i}{2}

乘以:

1 i = i

= \frac{i}{2}

= \frac{1}{2} i

化简

- - \frac{1}{4} : - i \frac{1}{2}

- - \frac{1}{4}

化简

- \frac{1}{4} : i \frac{1}{2}

- \frac{1}{4}

使用根式运算法则:

- a = - 1 a

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

= - 1 \frac{1}{4}

使用虚数运算法则:

- 1 = i

= i \frac{1}{4}

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

= i \frac{1}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= i \frac{1}{2}

= - i \frac{1}{2}

使用法则

1 = 1

= - \frac{1}{2} i

u = i \frac{1}{2}, u = - i \frac{1}{2}

u = sin (x)

代回

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

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

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

无解

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

无解

sin (x) = - i \frac{1}{2} :

无解

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

无解

合并所有解

无解

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

作图

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