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受欢迎的三角函数 >

(2cos(x)-1)*(2sin(x)+1)=3-4sin^2(x)

解答

(2 cos (x) - 1) \cdot (2 sin (x) + 1) = 3 - 4 sin^{2} (x)

解答

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

+1

度数

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

(2 cos (x) - 1) (2 sin (x) + 1) = 3 - 4 sin^{2} (x)

两边减去

3 - 4 sin^{2} (x)

4 sin^{2} (x) + 4 cos (x) sin (x) + 2 cos (x) - 2 sin (x) - 4 = 0

使用三角恒等式改写

- 4 + 2 cos (x) - 2 sin (x) + 4 sin^{2} (x) + 4 cos (x) sin (x)

使用毕达哥拉斯恒等式:

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

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

= - 4 + 2 cos (x) - 2 sin (x) + 4 (1 - cos^{2} (x)) + 4 cos (x) sin (x)

化简

- 4 + 2 cos (x) - 2 sin (x) + 4 (1 - cos^{2} (x)) + 4 cos (x) sin (x) : 2 cos (x) + 4 cos (x) sin (x) - 4 cos^{2} (x) - 2 sin (x)

- 4 + 2 cos (x) - 2 sin (x) + 4 (1 - cos^{2} (x)) + 4 cos (x) sin (x)

乘开

4 (1 - cos^{2} (x)) : 4 - 4 cos^{2} (x)

4 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (x)

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

数字相乘:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (x)

= - 4 + 2 cos (x) - 2 sin (x) + 4 - 4 cos^{2} (x) + 4 cos (x) sin (x)

化简

- 4 + 2 cos (x) - 2 sin (x) + 4 - 4 cos^{2} (x) + 4 cos (x) sin (x) : 2 cos (x) + 4 cos (x) sin (x) - 4 cos^{2} (x) - 2 sin (x)

- 4 + 2 cos (x) - 2 sin (x) + 4 - 4 cos^{2} (x) + 4 cos (x) sin (x)

对同类项分组

= 2 cos (x) - 2 sin (x) - 4 cos^{2} (x) + 4 cos (x) sin (x) - 4 + 4

- 4 + 4 = 0

= 2 cos (x) + 4 cos (x) sin (x) - 4 cos^{2} (x) - 2 sin (x)

= 2 cos (x) + 4 cos (x) sin (x) - 4 cos^{2} (x) - 2 sin (x)

= 2 cos (x) + 4 cos (x) sin (x) - 4 cos^{2} (x) - 2 sin (x)

2 cos (x) - 2 sin (x) - 4 cos^{2} (x) + 4 cos (x) sin (x) = 0

分解

2 cos (x) - 2 sin (x) - 4 cos^{2} (x) + 4 cos (x) sin (x) : - 2 (2 cos (x) - 1) (cos (x) - sin (x))

2 cos (x) - 2 sin (x) - 4 cos^{2} (x) + 4 cos (x) sin (x)

将

4

改写为

2 \cdot 2

将

- 4

改写为

2 \cdot 2

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

因式分解出通项

2

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

分解

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

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

分解

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

cos (x) - 2 cos^{2} (x)

使用指数法则:

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

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

= cos (x) - 2 cos (x) cos (x)

因式分解出通项

cos (x)

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

分解

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

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

因式分解出通项

sin (x)

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

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

改写为

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

因式分解出通项

(1 - 2 cos (x))

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

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

分解

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

1 - 2 cos (x)

因式分解出通项

- 1

= - (2 cos (x) - 1)

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

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

分别求解每个部分

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

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

2 cos (x) - 1 = 0

将

1

到右边

2 cos (x) - 1 = 0

两边加上

1

2 cos (x) - 1 + 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

cos (x) - sin (x) = 0 : x = \frac{π}{4} + πn

cos (x) - sin (x) = 0

使用三角恒等式改写

cos (x) - sin (x) = 0

在两边除以

cos (x), cos (x) \neq = 0

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

化简

1 - \frac{sin ( x )}{cos ( x )} = 0

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 - tan (x) = 0

1 - tan (x) = 0

将

1

到右边

1 - tan (x) = 0

两边减去

1

1 - tan (x) - 1 = 0 - 1

化简

- tan (x) = - 1

- tan (x) = - 1

两边除以

- 1

- tan (x) = - 1

两边除以

- 1

\frac{- tan ( x )}{- 1} = \frac{- 1}{- 1}

化简

tan (x) = 1

tan (x) = 1

tan (x) = 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

合并所有解

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

作图

流行的例子

r=(2a)/((1+cos(x)))

cos^3(x)+cos^2(x)+cos(x)+1=0

(cos^4(x))/3 =sin^2(x)

sin^3(x)cos(x)-sin^2(x)=0

sin^2(x)+sin^4(x)=0