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受欢迎的三角函数 >

(2cos(x)-sin(x))(1+sin(x))=cos^2(x)

解答

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

解答

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

+1

度数

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

cos^{2} (x)

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

- (1 - sin^{2} (x)) : - 1 + sin^{2} (x)

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

打开括号

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

使用加减运算法则

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

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

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

乘开

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

(- sin (x) + 2 cos (x)) (1 + sin (x))

使用 FOIL 方法:

(a + b) (c + d) = a c + a d + b c + b d

a = - sin (x), b = 2 cos (x), c = 1, d = sin (x)

= (- sin (x)) \cdot 1 + (- sin (x)) sin (x) + 2 cos (x) \cdot 1 + 2 cos (x) sin (x)

使用加减运算法则

+ (- a) = - a

= - 1 \cdot sin (x) - sin (x) sin (x) + 2 \cdot 1 \cdot cos (x) + 2 cos (x) sin (x)

化简

- 1 \cdot sin (x) - sin (x) sin (x) + 2 \cdot 1 \cdot cos (x) + 2 cos (x) sin (x) : - sin (x) - sin^{2} (x) + 2 cos (x) + 2 cos (x) sin (x)

- 1 \cdot sin (x) - sin (x) sin (x) + 2 \cdot 1 \cdot cos (x) + 2 cos (x) sin (x)

1 \cdot sin (x) = sin (x)

1 \cdot sin (x)

乘以:

1 \cdot sin (x) = sin (x)

= sin (x)

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

sin (x) sin (x)

使用指数法则:

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

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

= sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= sin^{2} (x)

2 \cdot 1 \cdot cos (x) = 2 cos (x)

2 \cdot 1 \cdot cos (x)

数字相乘:

2 \cdot 1 = 2

= 2 cos (x)

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

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

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

化简

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

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

对同类项分组

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

同类项相加:

sin^{2} (x) - sin^{2} (x) = 0

= - sin (x) + 2 cos (x) + 2 cos (x) sin (x) - 1

= - sin (x) + 2 cos (x) + 2 cos (x) sin (x) - 1

= - sin (x) + 2 cos (x) + 2 cos (x) sin (x) - 1

使用基本三角恒等式:

sin (x) = \frac{1}{csc ( x )}

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

化简

- 1 - \frac{1}{csc ( x )} + 2 cos (x) + 2 cos (x) \frac{1}{csc ( x )} : - 1 + \frac{- 1 + 2 cos ( x )}{csc ( x )} + 2 cos (x)

- 1 - \frac{1}{csc ( x )} + 2 cos (x) + 2 cos (x) \frac{1}{csc ( x )}

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

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

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

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

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

合并分式

- \frac{1}{csc ( x )} + \frac{2 cos ( x )}{csc ( x )} : \frac{- 1 + 2 cos ( x )}{csc ( x )}

使用法则

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

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

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

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

使用基本三角恒等式:

\frac{1}{csc ( x )} = sin (x)

= - 1 + (- 1 + 2 cos (x)) sin (x) + 2 cos (x)

- 1 + (- 1 + 2 cos (x)) sin (x) + 2 cos (x) = 0

分解

- 1 + (- 1 + 2 cos (x)) sin (x) + 2 cos (x) : (- 1 + 2 cos (x)) (sin (x) + 1)

- 1 + (- 1 + 2 cos (x)) sin (x) + 2 cos (x)

改写为

= (- 1 + 2 cos (x)) sin (x) + 1 \cdot (- 1 + 2 cos (x))

因式分解出通项

(- 1 + 2 cos (x))

= (- 1 + 2 cos (x)) (sin (x) + 1)

(- 1 + 2 cos (x)) (sin (x) + 1) = 0

分别求解每个部分

- 1 + 2 cos (x) = 0 or sin (x) + 1 = 0

- 1 + 2 cos (x) = 0 : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

- 1 + 2 cos (x) = 0

将

1

到右边

- 1 + 2 cos (x) = 0

两边加上

1

- 1 + 2 cos (x) + 1 = 0 + 1

化简

2 cos (x) = 1

2 cos (x) = 1

两边除以

2

2 cos (x) = 1

两边除以

2

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

化简

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

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

sin (x) + 1 = 0 : x = \frac{3 π}{2} + 2 πn

sin (x) + 1 = 0

将

1

到右边

sin (x) + 1 = 0

两边减去

1

sin (x) + 1 - 1 = 0 - 1

化简

sin (x) = - 1

sin (x) = - 1

sin (x) = - 1

的通解

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

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

合并所有解

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

作图

流行的例子

sin(6x)cos(9x)-cos(6x)sin(9x)=-0.4

sin (6 x) cos (9 x) - cos (6 x) sin (9 x) = - 0.4

sin^2(x)=2tan^2(x)

sin^{2} (x) = 2 tan^{2} (x)

cos^3(x)=cos(2x+x)

cos^{3} (x) = cos (2 x + x)

solvefor x,(d^2-3d+2)y=sin(e^x)

so l v e f or x, (d^{2} - 3 d + 2) y = sin (e^{x})

cos(6x)+cos(2x)=0

cos (6 x) + cos (2 x) = 0