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受欢迎的三角函数 >

(cos^2(a)-3cos(a)+2)/(sin^2(a))=1

解答

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} = 1

解答

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

+1

度数

a = 6 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n

求解步骤

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} = 1

两边减去

1

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} - 1 = 0

化简

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} - 1 : \frac{cos ^{2} ( a ) - 3 cos ( a ) + 2 - sin ^{2} ( a )}{sin ^{2} ( a )}

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} - 1

将项转换为分式:

1 = \frac{1 sin ^{2} ( a )}{sin ^{2} ( a )}

= \frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} - \frac{1 \cdot sin ^{2} ( a )}{sin ^{2} ( a )}

因为分母相等，所以合并分式:

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

= \frac{cos ^{2} ( a ) - 3 cos ( a ) + 2 - 1 \cdot sin ^{2} ( a )}{sin ^{2} ( a )}

乘以:

1 \cdot sin^{2} (a) = sin^{2} (a)

= \frac{cos ^{2} ( a ) - 3 cos ( a ) + 2 - sin ^{2} ( a )}{sin ^{2} ( a )}

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2 - sin ^{2} ( a )}{sin ^{2} ( a )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cos^{2} (a) - 3 cos (a) + 2 - sin^{2} (a) = 0

使用三角恒等式改写

2 + cos^{2} (a) - sin^{2} (a) - 3 cos (a)

使用毕达哥拉斯恒等式:

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

sin^{2} (x) = 1 - cos^{2} (x)

= 2 + cos^{2} (a) - (1 - cos^{2} (a)) - 3 cos (a)

化简

2 + cos^{2} (a) - (1 - cos^{2} (a)) - 3 cos (a) : 2 cos^{2} (a) - 3 cos (a) + 1

2 + cos^{2} (a) - (1 - cos^{2} (a)) - 3 cos (a)

- (1 - cos^{2} (a)) : - 1 + cos^{2} (a)

- (1 - cos^{2} (a))

打开括号

= - (1) - (- cos^{2} (a))

使用加减运算法则

- (- a) = a, - (a) = - a

= - 1 + cos^{2} (a)

= 2 + cos^{2} (a) - 1 + cos^{2} (a) - 3 cos (a)

化简

2 + cos^{2} (a) - 1 + cos^{2} (a) - 3 cos (a) : 2 cos^{2} (a) - 3 cos (a) + 1

2 + cos^{2} (a) - 1 + cos^{2} (a) - 3 cos (a)

对同类项分组

= cos^{2} (a) + cos^{2} (a) - 3 cos (a) + 2 - 1

同类项相加:

cos^{2} (a) + cos^{2} (a) = 2 cos^{2} (a)

= 2 cos^{2} (a) - 3 cos (a) + 2 - 1

数字相加/相减:

2 - 1 = 1

= 2 cos^{2} (a) - 3 cos (a) + 1

= 2 cos^{2} (a) - 3 cos (a) + 1

= 2 cos^{2} (a) - 3 cos (a) + 1

1 + 2 cos^{2} (a) - 3 cos (a) = 0

用替代法求解

1 + 2 cos^{2} (a) - 3 cos (a) = 0

令

: cos (a) = u

1 + 2 u^{2} - 3 u = 0

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

1 + 2 u^{2} - 3 u = 0

改写成标准形式

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

2 u^{2} - 3 u + 1 = 0

使用求根公式求解

2 u^{2} - 3 u + 1 = 0

二次方程求根公式:

若

a = 2, b = - 3, c = 1

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

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

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

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

使用指数法则:

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

若

n

是偶数

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

= 3^{2} - 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数字相减:

9 - 8 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

3 + 1 = 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

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

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

使用法则

- (- a) = a

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

数字相减:

3 - 1 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

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

u = cos (a)

代回

cos (a) = 1, cos (a) = \frac{1}{2}

cos (a) = 1, cos (a) = \frac{1}{2}

cos (a) = 1 : a = 2 πn

cos (a) = 1

cos (a) = 1

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}

a = 0 + 2 πn

a = 0 + 2 πn

解

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn

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

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

cos (a) = \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}

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

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

合并所有解

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

因为方程对以下值无定义:

2 πn

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

作图

流行的例子

(sin(x)-(sqrt(2)))/2 =0

\frac{sin ( x ) - ( 2 )}{2} = 0

cos(2x)=5-6cos^2(x)

cos (2 x) = 5 - 6 cos^{2} (x)

cos^{4} (x) = 0.37

sin^2(x)=((sqrt(3)))/2

sin^{2} (x) = \frac{( 3 )}{2}

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

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