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受欢迎的三角函数 >

4sin^2(x)cot(x)-cot(x)=0,0<= x<= 2pi

解答

4 sin^{2} (x) cot (x) - cot (x) = 0, 0 \leq x \leq 2 π

解答

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = \frac{π}{6}, x = \frac{5 π}{6}

+1

度数

x = 9 0^{\circ}, x = 27 0^{\circ}, x = 21 0^{\circ}, x = 33 0^{\circ}, x = 3 0^{\circ}, x = 15 0^{\circ}

求解步骤

4 sin^{2} (x) cot (x) - cot (x) = 0, 0 \leq x \leq 2 π

分解

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

4 sin^{2} (x) cot (x) - cot (x)

因式分解出通项

cot (x)

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

分解

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

4 sin^{2} (x) - 1

将

4 sin^{2} (x) - 1

改写为

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

4 sin^{2} (x) - 1

将

4

改写为

2^{2}

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

将

1

改写为

1^{2}

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

使用指数法则:

a^{m} b^{m} = (ab)^{m}

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

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

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

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

分别求解每个部分

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

cot (x) = 0, 0 \leq x \leq 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cot (x) = 0, 0 \leq x \leq 2 π

cot (x) = 0

的通解

cot (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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

在

0 \leq x \leq 2 π

范围内的解

x = \frac{π}{2}, x = \frac{3 π}{2}

2 sin (x) + 1 = 0, 0 \leq x \leq 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

2 sin (x) + 1 = 0, 0 \leq x \leq 2 π

将

1

到右边

2 sin (x) + 1 = 0

两边减去

1

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

化简

2 sin (x) = - 1

2 sin (x) = - 1

两边除以

2

2 sin (x) = - 1

两边除以

2

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

化简

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

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

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

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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = \frac{7 π}{6}, x = \frac{11 π}{6}

2 sin (x) - 1 = 0, 0 \leq x \leq 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

2 sin (x) - 1 = 0, 0 \leq x \leq 2 π

将

1

到右边

2 sin (x) - 1 = 0

两边加上

1

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

化简

2 sin (x) = 1

2 sin (x) = 1

两边除以

2

2 sin (x) = 1

两边除以

2

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

化简

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

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

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

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{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = \frac{π}{6}, x = \frac{5 π}{6}

合并所有解

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = \frac{π}{6}, x = \frac{5 π}{6}

作图

流行的例子

14-sin(θ)=cos(2θ)

14 - sin (θ) = cos (2 θ)

csc (θ) = - \frac{5}{13}

cos (z) = 3

4cos(2x)=1-3cos(x)

4 cos (2 x) = 1 - 3 cos (x)

0=32pisec(x^2)tan(x)

0 = 32 π sec (x^{2}) tan (x)