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受欢迎的三角函数 >

-a^2c_{1}cos(ax)-a^2c_{2}sin(ax)=0

解答

- a^{2} c_{1} cos (a x) - a^{2} c_{2} sin (a x) = 0

解答

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

求解步骤

- a^{2} c_{1} cos (a x) - a^{2} c_{2} sin (a x) = 0

使用三角恒等式改写

- a^{2} c_{1} cos (a x) - a^{2} c_{2} sin (a x) = 0

在两边除以

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

\frac{- a ^{2} c _{1} cos ( a x ) - a ^{2} c _{2} sin ( a x )}{cos ( a x )} = \frac{0}{cos ( a x )}

化简

- a^{2} c_{1} - \frac{a ^{2} c _{2} sin ( a x )}{cos ( a x )} = 0

使用基本三角恒等式:

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

- a^{2} c_{1} - a^{2} c_{2} tan (a x) = 0

- a^{2} c_{1} - a^{2} c_{2} tan (a x) = 0

将

a^{2} c_{1}

到右边

- a^{2} c_{1} - a^{2} c_{2} tan (a x) = 0

两边加上

a^{2} c_{1}

- a^{2} c_{1} - a^{2} c_{2} tan (a x) + a^{2} c_{1} = 0 + a^{2} c_{1}

化简

- a^{2} c_{2} tan (a x) = a^{2} c_{1}

- a^{2} c_{2} tan (a x) = a^{2} c_{1}

两边除以

- a^{2} c_{2}; a \neq = 0

- a^{2} c_{2} tan (a x) = a^{2} c_{1}

两边除以

- a^{2} c_{2}; a \neq = 0

\frac{- a ^{2} c _{2} tan ( a x )}{- a ^{2} c _{2}} = \frac{a ^{2} c _{1}}{- a ^{2} c _{2}}; a \neq = 0

化简

\frac{- a ^{2} c _{2} tan ( a x )}{- a ^{2} c _{2}} = \frac{a ^{2} c _{1}}{- a ^{2} c _{2}}

化简

\frac{- a ^{2} c _{2} tan ( a x )}{- a ^{2} c _{2}} : tan (a x)

\frac{- a ^{2} c _{2} tan ( a x )}{- a ^{2} c _{2}}

使用分式法则:

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

= \frac{a ^{2} c _{2} tan ( a x )}{a ^{2} c _{2}}

约分:

a^{2}

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

约分:

c_{2}

= tan (a x)

化简

\frac{a ^{2} c _{1}}{- a ^{2} c _{2}} : - \frac{c _{1}}{c _{2}}

\frac{a ^{2} c _{1}}{- a ^{2} c _{2}}

使用分式法则:

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

= - \frac{a ^{2} c _{1}}{a ^{2} c _{2}}

约分:

a^{2}

= - \frac{c _{1}}{c _{2}}

tan (a x) = - \frac{c _{1}}{c _{2}}; a \neq = 0

tan (a x) = - \frac{c _{1}}{c _{2}}; a \neq = 0

tan (a x) = - \frac{c _{1}}{c _{2}}; a \neq = 0

使用反三角函数性质

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

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

的通解

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

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

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

解

a x = arctan (- \frac{c _{1}}{c _{2}}) + πn : x = \frac{arctan ( - \frac{c _{1}}{c _{2}} )}{a} + \frac{πn}{a}; a \neq = 0

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

两边除以

a; a \neq = 0

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

两边除以

a; a \neq = 0

\frac{a x}{a} = \frac{arctan ( - \frac{c _{1}}{c _{2}} )}{a} + \frac{πn}{a}; a \neq = 0

化简

x = \frac{arctan ( - \frac{c _{1}}{c _{2}} )}{a} + \frac{πn}{a}; a \neq = 0

x = \frac{arctan ( - \frac{c _{1}}{c _{2}} )}{a} + \frac{πn}{a}; a \neq = 0

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

作图

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