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受欢迎的三角函数 >

solvefor x,cos^2(x)z^2-2cos^2(x)z+1=0

解答

求解

x, cos^{2} (x) \cdot z^{2} - 2 cos^{2} (x) \cdot z + 1 = 0

解答

x \in R 无解

求解步骤

cos^{2} (x) z^{2} - 2 cos^{2} (x) z + 1 = 0

用替代法求解

cos^{2} (x) z^{2} - 2 cos^{2} (x) z + 1 = 0

令

: cos (x) = u

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

u^{2} z^{2} - 2 u^{2} z + 1 = 0 : u = i \frac{1}{z ^{2} - 2 z}, u = - i \frac{1}{z ^{2} - 2 z}; z \neq = 0, z \neq = 2

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

将

1

到右边

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

两边减去

1

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

化简

u^{2} z^{2} - 2 u^{2} z = - 1

u^{2} z^{2} - 2 u^{2} z = - 1

分解

u^{2} z^{2} - 2 u^{2} z : z u^{2} (z - 2)

u^{2} z^{2} - 2 u^{2} z

使用指数法则:

a^{b + c} = a^{b} a^{c}

z^{2} = zz

= u^{2} zz - 2 u^{2} z

因式分解出通项

u^{2} z

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

z u^{2} (z - 2) = - 1

两边除以

z (z - 2); z \neq = 0, z \neq = 2

z u^{2} (z - 2) = - 1

两边除以

z (z - 2); z \neq = 0, z \neq = 2

\frac{z u ^{2} ( z - 2 )}{z ( z - 2 )} = \frac{- 1}{z ( z - 2 )}; z \neq = 0, z \neq = 2

化简

u^{2} = - \frac{1}{z ( z - 2 )}; z \neq = 0, z \neq = 2

u^{2} = - \frac{1}{z ( z - 2 )}; z \neq = 0, z \neq = 2

对于

x^{2} = f (a)

解为

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

u = - \frac{1}{z ( z - 2 )}, u = - - \frac{1}{z ( z - 2 )}; z \neq = 0, z \neq = 2

化简

- \frac{1}{z ( z - 2 )} : i \frac{1}{z ^{2} - 2 z}

- \frac{1}{z ( z - 2 )}

使用根式运算法则:

- a = - 1 a,

假定

a \geq 0

- \frac{1}{z ( z - 2 )} = - 1 \frac{1}{z ( z - 2 )}

= - 1 \frac{1}{z ( z - 2 )}

使用虚数运算法则:

- 1 = i

= i \frac{1}{z ( z - 2 )}

将

i \frac{1}{z ( z - 2 )}

改写成标准复数形式:

\frac{1}{z ^{2} - 2 z} i

i \frac{1}{z ( z - 2 )}

\frac{1}{z ( z - 2 )} = \frac{1}{z ^{2} - 2 z}

\frac{1}{z ( z - 2 )}

\frac{1}{z ( z - 2 )} = \frac{1}{z ^{2} - 2 z}

\frac{1}{z ( z - 2 )}

乘开

z (z - 2) : z^{2} - 2 z

z (z - 2)

使用分配律:

a (b - c) = ab - a c

a = z, b = z, c = 2

= zz - z \cdot 2

= zz - 2 z

zz = z^{2}

zz

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

zz = z^{1 + 1}

= z^{1 + 1}

数字相加:

1 + 1 = 2

= z^{2}

= z^{2} - 2 z

= \frac{1}{z ^{2} - 2 z}

= \frac{1}{z ^{2} - 2 z}

= i \frac{1}{z ^{2} - 2 z}

= \frac{1}{z ^{2} - 2 z} i

化简

- - \frac{1}{z ( z - 2 )} : - i \frac{1}{z ^{2} - 2 z}

- - \frac{1}{z ( z - 2 )}

化简

- \frac{1}{z ( z - 2 )} : i \frac{1}{z ( z - 2 )}

- \frac{1}{z ( z - 2 )}

使用根式运算法则:

- a = - 1 a,

假定

a \geq 0

- \frac{1}{z ( z - 2 )} = - 1 \frac{1}{z ( z - 2 )}

= - 1 \frac{1}{z ( z - 2 )}

使用虚数运算法则:

- 1 = i

= i \frac{1}{z ( z - 2 )}

= - i \frac{1}{z ( z - 2 )}

将

- i \frac{1}{z ( z - 2 )}

改写成标准复数形式:

- \frac{1}{z ^{2} - 2 z} i

- i \frac{1}{z ( z - 2 )}

\frac{1}{z ( z - 2 )} = \frac{1}{z ^{2} - 2 z}

\frac{1}{z ( z - 2 )}

\frac{1}{z ( z - 2 )} = \frac{1}{z ^{2} - 2 z}

\frac{1}{z ( z - 2 )}

乘开

z (z - 2) : z^{2} - 2 z

z (z - 2)

使用分配律:

a (b - c) = ab - a c

a = z, b = z, c = 2

= zz - z \cdot 2

= zz - 2 z

zz = z^{2}

zz

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

zz = z^{1 + 1}

= z^{1 + 1}

数字相加:

1 + 1 = 2

= z^{2}

= z^{2} - 2 z

= \frac{1}{z ^{2} - 2 z}

= \frac{1}{z ^{2} - 2 z}

= - i \frac{1}{z ^{2} - 2 z}

= - \frac{1}{z ^{2} - 2 z} i

u = i \frac{1}{z ^{2} - 2 z}, u = - i \frac{1}{z ^{2} - 2 z}; z \neq = 0, z \neq = 2

u = cos (x)

代回

cos (x) = i \frac{1}{z ^{2} - 2 z}, cos (x) = - i \frac{1}{z ^{2} - 2 z}; z \neq = 0, z \neq = 2

cos (x) = i \frac{1}{z ^{2} - 2 z}, cos (x) = - i \frac{1}{z ^{2} - 2 z}; z \neq = 0, z \neq = 2

cos (x) = i \frac{1}{z ^{2} - 2 z} :

无解

cos (x) = i \frac{1}{z ^{2} - 2 z}

无解

cos (x) = - i \frac{1}{z ^{2} - 2 z} :

无解

cos (x) = - i \frac{1}{z ^{2} - 2 z}

无解

合并所有解

x \in R 无解

作图

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