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受欢迎的三角函数 >

2cos^2(t)tan(t)-tan(t)=0

解答

2 cos^{2} (t) tan (t) - tan (t) = 0

解答

t = πn, t = \frac{3 π}{4} + 2 πn, t = \frac{5 π}{4} + 2 πn, t = \frac{π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

+1

度数

t = 0^{\circ} + 18 0^{\circ} n, t = 13 5^{\circ} + 36 0^{\circ} n, t = 22 5^{\circ} + 36 0^{\circ} n, t = 4 5^{\circ} + 36 0^{\circ} n, t = 31 5^{\circ} + 36 0^{\circ} n

求解步骤

2 cos^{2} (t) tan (t) - tan (t) = 0

分解

2 cos^{2} (t) tan (t) - tan (t) : tan (t) (2 cos (t) + 1) (2 cos (t) - 1)

2 cos^{2} (t) tan (t) - tan (t)

因式分解出通项

tan (t)

= tan (t) (2 cos^{2} (t) - 1)

分解

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

2 cos^{2} (t) - 1

将

2 cos^{2} (t) - 1

改写为

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

2 cos^{2} (t) - 1

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

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

将

1

改写为

1^{2}

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

使用指数法则:

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

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

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

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

使用平方差公式:

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

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

= (2 cos (t) + 1) (2 cos (t) - 1)

= tan (t) (2 cos (t) + 1) (2 cos (t) - 1)

tan (t) (2 cos (t) + 1) (2 cos (t) - 1) = 0

分别求解每个部分

tan (t) = 0 or 2 cos (t) + 1 = 0 or 2 cos (t) - 1 = 0

tan (t) = 0 : t = πn

tan (t) = 0

tan (t) = 0

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}

t = 0 + πn

t = 0 + πn

解

t = 0 + πn : t = πn

t = 0 + πn

0 + πn = πn

t = πn

t = πn

2 cos (t) + 1 = 0 : t = \frac{3 π}{4} + 2 πn, t = \frac{5 π}{4} + 2 πn

2 cos (t) + 1 = 0

将

1

到右边

2 cos (t) + 1 = 0

两边减去

1

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

化简

2 cos (t) = - 1

2 cos (t) = - 1

两边除以

2

2 cos (t) = - 1

两边除以

2

\frac{2 cos ( t )}{2} = \frac{- 1}{2}

化简

\frac{2 cos ( t )}{2} = \frac{- 1}{2}

化简

\frac{2 cos ( t )}{2} : cos (t)

\frac{2 cos ( t )}{2}

约分:

2

= cos (t)

化简

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

cos (t) = - \frac{2}{2}

cos (t) = - \frac{2}{2}

cos (t) = - \frac{2}{2}

cos (t) = - \frac{2}{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}

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

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

2 cos (t) - 1 = 0 : t = \frac{π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

2 cos (t) - 1 = 0

将

1

到右边

2 cos (t) - 1 = 0

两边加上

1

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

化简

2 cos (t) = 1

2 cos (t) = 1

两边除以

2

2 cos (t) = 1

两边除以

2

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

化简

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

化简

\frac{2 cos ( t )}{2} : cos (t)

\frac{2 cos ( t )}{2}

约分:

2

= cos (t)

化简

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

cos (t) = \frac{2}{2}

cos (t) = \frac{2}{2}

cos (t) = \frac{2}{2}

cos (t) = \frac{2}{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}

t = \frac{π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

t = \frac{π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

合并所有解

t = πn, t = \frac{3 π}{4} + 2 πn, t = \frac{5 π}{4} + 2 πn, t = \frac{π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

作图

流行的例子

3 = 2 - 2 sin (x)

(cos(x))/(csc(x))+(sin(x))/(sec(x))=1

\frac{cos ( x )}{csc ( x )} + \frac{sin ( x )}{sec ( x )} = 1

cot (θ) = - 8

solvefor t,0=(1.8)cos(2pift)

so l v e f or t, 0 = (1.8) cos (2 π f t)

2cot(x)cos(x)+2cos(x)=cot(x)+1

2 cot (x) cos (x) + 2 cos (x) = cot (x) + 1