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受欢迎的三角函数 >

2sin(2x)+cos(2x)=0

解答

2 sin (2 x) + cos (2 x) = 0

解答

x = - \frac{0.46364 \dots}{2} + \frac{πn}{2}

+1

度数

x = - 13.28252 \dots^{\circ} + 9 0^{\circ} n

求解步骤

2 sin (2 x) + cos (2 x) = 0

使用三角恒等式改写

2 sin (2 x) + cos (2 x) = 0

在两边除以

cos (2 x), cos (2 x) \neq = 0

\frac{2 sin ( 2 x ) + cos ( 2 x )}{cos ( 2 x )} = \frac{0}{cos ( 2 x )}

化简

\frac{2 sin ( 2 x )}{cos ( 2 x )} + 1 = 0

使用基本三角恒等式:

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

2 tan (2 x) + 1 = 0

2 tan (2 x) + 1 = 0

将

1

到右边

2 tan (2 x) + 1 = 0

两边减去

1

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

化简

2 tan (2 x) = - 1

2 tan (2 x) = - 1

两边除以

2

2 tan (2 x) = - 1

两边除以

2

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

化简

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

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

使用反三角函数性质

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

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

的通解

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

2 x = arctan (- \frac{1}{2}) + πn

2 x = arctan (- \frac{1}{2}) + πn

解

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

2 x = 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

2 x = - arctan (\frac{1}{2}) + πn

两边除以

2

2 x = - arctan (\frac{1}{2}) + πn

两边除以

2

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

化简

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

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

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

以小数形式表示解

x = - \frac{0.46364 \dots}{2} + \frac{πn}{2}

作图

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