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受欢迎的三角函数 >

-13asin(2x)-13bcos(2x)=0

解答

- 13 a \cdot sin (2 x) - 13 b \cdot cos (2 x) = 0

解答

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

求解步骤

- 13 a sin (2 x) - 13 b cos (2 x) = 0

使用三角恒等式改写

- 13 a sin (2 x) - 13 b cos (2 x) = 0

在两边除以

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

\frac{- 13 a sin ( 2 x ) - 13 b cos ( 2 x )}{cos ( 2 x )} = \frac{0}{cos ( 2 x )}

化简

- \frac{13 a sin ( 2 x )}{cos ( 2 x )} - 13 b = 0

使用基本三角恒等式:

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

- 13 a tan (2 x) - 13 b = 0

- 13 a tan (2 x) - 13 b = 0

将

13 b

到右边

- 13 a tan (2 x) - 13 b = 0

两边加上

13 b

- 13 a tan (2 x) - 13 b + 13 b = 0 + 13 b

化简

- 13 a tan (2 x) = 13 b

- 13 a tan (2 x) = 13 b

两边除以

- 13 a; a \neq = 0

- 13 a tan (2 x) = 13 b

两边除以

- 13 a; a \neq = 0

\frac{- 13 a tan ( 2 x )}{- 13 a} = \frac{13 b}{- 13 a}; a \neq = 0

化简

\frac{- 13 a tan ( 2 x )}{- 13 a} = \frac{13 b}{- 13 a}

化简

\frac{- 13 a tan ( 2 x )}{- 13 a} : tan (2 x)

\frac{- 13 a tan ( 2 x )}{- 13 a}

使用分式法则:

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

= \frac{13 a tan ( 2 x )}{13 a}

数字相除:

\frac{13}{13} = 1

= \frac{a tan ( 2 x )}{a}

约分:

a

= tan (2 x)

化简

\frac{13 b}{- 13 a} : - \frac{b}{a}

\frac{13 b}{- 13 a}

使用分式法则:

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

= - \frac{13 b}{13 a}

数字相除:

\frac{13}{13} = 1

= - \frac{b}{a}

tan (2 x) = - \frac{b}{a}; a \neq = 0

tan (2 x) = - \frac{b}{a}; a \neq = 0

tan (2 x) = - \frac{b}{a}; a \neq = 0

使用反三角函数性质

tan (2 x) = - \frac{b}{a}

tan (2 x) = - \frac{b}{a}

的通解

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

2 x = arctan (- \frac{b}{a}) + πn

2 x = arctan (- \frac{b}{a}) + πn

解

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

2 x = arctan (- \frac{b}{a}) + πn

两边除以

2

2 x = arctan (- \frac{b}{a}) + πn

两边除以

2

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

化简

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

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

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

作图

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