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受欢迎的三角函数 >

solvefor x,0=4-1/(cos^2(x))

解答

求解

x, 0 = 4 - \frac{1}{cos ^{2} ( x )}

解答

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

+1

度数

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

求解步骤

0 = 4 - \frac{1}{cos ^{2} ( x )}

交换两边

4 - \frac{1}{cos ^{2} ( x )} = 0

用替代法求解

4 - \frac{1}{cos ^{2} ( x )} = 0

令

: cos (x) = u

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

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

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

在两边乘以

u^{2}

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

在两边乘以

u^{2}

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

化简

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

化简

- \frac{1}{u ^{2}} u^{2} : - 1

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

分式相乘:

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

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

约分:

u^{2}

= - 1

化简

0 \cdot u^{2} : 0

0 \cdot u^{2}

使用法则

0 \cdot a = 0

= 0

4 u^{2} - 1 = 0

4 u^{2} - 1 = 0

4 u^{2} - 1 = 0

解

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

4 u^{2} - 1 = 0

将

1

到右边

4 u^{2} - 1 = 0

两边加上

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}

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

\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}

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{1}{2}

使用法则

1 = 1

= \frac{1}{2}

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

- \frac{1}{4}

化简

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

\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}

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{1}{2}

= - \frac{1}{2}

使用法则

1 = 1

= - \frac{1}{2}

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

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

的分母，令其等于零

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

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

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

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

u = cos (x)

代回

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

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

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

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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

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

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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

合并所有解

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

作图

流行的例子

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csc^{2} (θ) = 2 cot (θ)

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sin (θ) = \frac{7}{25}, cos (\frac{θ}{2}), 0^{\circ} < θ < 9 0^{\circ}

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