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受欢迎的三角函数 >

solvefor x,d^2+3d+2y=4cos^2(x)

解答

求解

x, d^{2} + 3 d + 2 y = 4 cos^{2} (x)

解答

x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

求解步骤

d^{2} + 3 d + 2 y = 4 cos^{2} (x)

交换两边

4 cos^{2} (x) = d^{2} + 3 d + 2 y

用替代法求解

4 cos^{2} (x) = d^{2} + 3 d + 2 y

令

: cos (x) = u

4 u^{2} = d^{2} + 3 d + 2 y

4 u^{2} = d^{2} + 3 d + 2 y : u = \frac{d ^{2} + 3 d + 2 y}{2}, u = - \frac{d ^{2} + 3 d + 2 y}{2}

4 u^{2} = d^{2} + 3 d + 2 y

两边除以

4

4 u^{2} = d^{2} + 3 d + 2 y

两边除以

4

\frac{4 u ^{2}}{4} = \frac{d ^{2}}{4} + \frac{3 d}{4} + \frac{2 y}{4}

化简

\frac{4 u ^{2}}{4} = \frac{d ^{2}}{4} + \frac{3 d}{4} + \frac{2 y}{4}

化简

\frac{4 u ^{2}}{4} : u^{2}

\frac{4 u ^{2}}{4}

数字相除:

\frac{4}{4} = 1

= u^{2}

化简

\frac{d ^{2}}{4} + \frac{3 d}{4} + \frac{2 y}{4} : \frac{d ^{2} + 3 d + 2 y}{4}

\frac{d ^{2}}{4} + \frac{3 d}{4} + \frac{2 y}{4}

使用法则

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

= \frac{d ^{2} + 3 d + 2 y}{4}

u^{2} = \frac{d ^{2} + 3 d + 2 y}{4}

u^{2} = \frac{d ^{2} + 3 d + 2 y}{4}

u^{2} = \frac{d ^{2} + 3 d + 2 y}{4}

对于

x^{2} = f (a)

解为

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

u = \frac{d ^{2} + 3 d + 2 y}{4}, u = - \frac{d ^{2} + 3 d + 2 y}{4}

化简

\frac{d ^{2} + 3 d + 2 y}{4} : \frac{d ^{2} + 3 d + 2 y}{2}

\frac{d ^{2} + 3 d + 2 y}{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{d ^{2} + 3 d + 2 y}{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

= \frac{d ^{2} + 3 d + 2 y}{2}

化简

- \frac{d ^{2} + 3 d + 2 y}{4} : - \frac{d ^{2} + 3 d + 2 y}{2}

- \frac{d ^{2} + 3 d + 2 y}{4}

化简

\frac{d ^{2} + 3 d + 2 y}{4} : \frac{d ^{2} + 3 d + 2 y}{2}

\frac{d ^{2} + 3 d + 2 y}{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{d ^{2} + 3 d + 2 y}{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

= \frac{d ^{2} + 3 d + 2 y}{2}

= - \frac{2 y + d ^{2} + 3 d}{2}

= - \frac{d ^{2} + 3 d + 2 y}{2}

u = \frac{d ^{2} + 3 d + 2 y}{2}, u = - \frac{d ^{2} + 3 d + 2 y}{2}

u = cos (x)

代回

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2}, cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2}

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2}, cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2}

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2} : x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2}

使用反三角函数性质

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2}

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2} : x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2}

使用反三角函数性质

cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2}

cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

合并所有解

x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

作图

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