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受欢迎的三角函数 >

2cot^2(t)sin(t)-cot^2(t)=0

解答

2 cot^{2} (t) sin (t) - cot^{2} (t) = 0

解答

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

+1

度数

t = 9 0^{\circ} + 18 0^{\circ} n, t = 3 0^{\circ} + 36 0^{\circ} n, t = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

2 cot^{2} (t) sin (t) - cot^{2} (t) = 0

分解

2 cot^{2} (t) sin (t) - cot^{2} (t) : cot^{2} (t) (2 sin (t) - 1)

2 cot^{2} (t) sin (t) - cot^{2} (t)

因式分解出通项

cot^{2} (t)

= cot^{2} (t) (2 sin (t) - 1)

cot^{2} (t) (2 sin (t) - 1) = 0

分别求解每个部分

cot^{2} (t) = 0 or 2 sin (t) - 1 = 0

cot^{2} (t) = 0 : t = \frac{π}{2} + πn

cot^{2} (t) = 0

使用法则

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

cot (t) = 0

cot (t) = 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

t = \frac{π}{2} + πn

t = \frac{π}{2} + πn

2 sin (t) - 1 = 0 : t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

2 sin (t) - 1 = 0

将

1

到右边

2 sin (t) - 1 = 0

两边加上

1

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

化简

2 sin (t) = 1

2 sin (t) = 1

两边除以

2

2 sin (t) = 1

两边除以

2

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

化简

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

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

sin (t) = \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}

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

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

合并所有解

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

作图

流行的例子

7 cos (x) = - 1

2sin^2(θ)+cos(θ)-2=-1

2 sin^{2} (θ) + cos (θ) - 2 = - 1

sin^2(x)-6cos(x)+6=0

sin^{2} (x) - 6 cos (x) + 6 = 0

cos(x)=(14.3)/(19)

cos (x) = \frac{14.3}{19}

- 12 sin (x) = 0