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受欢迎的三角函数 >

3cos(20t)+6sin(20t)=0

解答

3 cos (20 t) + 6 sin (20 t) = 0

解答

t = - \frac{0.46364 \dots}{20} + \frac{πn}{20}

+1

度数

t = - 1.32825 \dots^{\circ} + 9^{\circ} n

求解步骤

3 cos (20 t) + 6 sin (20 t) = 0

使用三角恒等式改写

3 cos (20 t) + 6 sin (20 t) = 0

在两边除以

cos (20 t), cos (20 t) \neq = 0

\frac{3 cos ( 20 t ) + 6 sin ( 20 t )}{cos ( 20 t )} = \frac{0}{cos ( 20 t )}

化简

3 + \frac{6 sin ( 20 t )}{cos ( 20 t )} = 0

使用基本三角恒等式:

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

3 + 6 tan (20 t) = 0

3 + 6 tan (20 t) = 0

将

3

到右边

3 + 6 tan (20 t) = 0

两边减去

3

3 + 6 tan (20 t) - 3 = 0 - 3

化简

6 tan (20 t) = - 3

6 tan (20 t) = - 3

两边除以

6

6 tan (20 t) = - 3

两边除以

6

\frac{6 tan ( 20 t )}{6} = \frac{- 3}{6}

化简

\frac{6 tan ( 20 t )}{6} = \frac{- 3}{6}

化简

\frac{6 tan ( 20 t )}{6} : tan (20 t)

\frac{6 tan ( 20 t )}{6}

数字相除:

\frac{6}{6} = 1

= tan (20 t)

化简

\frac{- 3}{6} : - \frac{1}{2}

\frac{- 3}{6}

使用分式法则:

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

= - \frac{3}{6}

约分:

3

= - \frac{1}{2}

tan (20 t) = - \frac{1}{2}

tan (20 t) = - \frac{1}{2}

tan (20 t) = - \frac{1}{2}

使用反三角函数性质

tan (20 t) = - \frac{1}{2}

tan (20 t) = - \frac{1}{2}

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

20 t = arctan (- \frac{1}{2}) + πn

20 t = arctan (- \frac{1}{2}) + πn

解

20 t = arctan (- \frac{1}{2}) + πn : t = - \frac{arctan ( \frac{1}{2} )}{20} + \frac{πn}{20}

20 t = arctan (- \frac{1}{2}) + πn

化简

arctan (- \frac{1}{2}) + πn : - arctan (\frac{1}{2}) + πn

arctan (- \frac{1}{2}) + πn

利用以下特性:

arctan (- x) = - arctan (x)

arctan (- \frac{1}{2}) = - arctan (\frac{1}{2})

= - arctan (\frac{1}{2}) + πn

20 t = - arctan (\frac{1}{2}) + πn

两边除以

20

20 t = - arctan (\frac{1}{2}) + πn

两边除以

20

\frac{20 t}{20} = - \frac{arctan ( \frac{1}{2} )}{20} + \frac{πn}{20}

化简

t = - \frac{arctan ( \frac{1}{2} )}{20} + \frac{πn}{20}

t = - \frac{arctan ( \frac{1}{2} )}{20} + \frac{πn}{20}

t = - \frac{arctan ( \frac{1}{2} )}{20} + \frac{πn}{20}

以小数形式表示解

t = - \frac{0.46364 \dots}{20} + \frac{πn}{20}

作图

流行的例子

tan (θ) = \frac{6}{2}

tan (θ) = \frac{6}{4}

2cot^2(x)+3csc(x)=0

2 cot^{2} (x) + 3 csc (x) = 0

solvefor t,f=2sin(kt)

so l v e f or t, f = 2 sin (k t)

cos(((2x-pi))/4)=1

cos (\frac{( 2 x - π )}{4}) = 1