What is the general solution for -ac_{1}sin(ax)+ac_{2}cos(ax)=0 ?

The general solution for -ac_{1}sin(ax)+ac_{2}cos(ax)=0 is x=(arctan((c_{2})/(c_{1)}))/a+(pin)/a

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Popular >

-ac_{1}sin(ax)+ac_{2}cos(ax)=0

Solution

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

Solution

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

Solution steps

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

Rewrite using trig identities

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

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

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

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

Move

a c_{2}

to the right side

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

Subtract

a c_{2}

from both sides

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

Simplify

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

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

Divide both sides by

- a c_{1}; a \neq = 0

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

Divide both sides by

- a c_{1}; a \neq = 0

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

Simplify

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

Simplify

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

a

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

Cancel the common factor:

c_{1}

= tan (a x)

Simplify

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

a

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Solve

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

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

Divide both sides by

a; a \neq = 0

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

Divide both sides by

a; a \neq = 0

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

Simplify

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

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

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

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Frequently Asked Questions (FAQ)

What is the general solution for -ac_{1}sin(ax)+ac_{2}cos(ax)=0 ?
The general solution for -ac_{1}sin(ax)+ac_{2}cos(ax)=0 is x=(arctan((c_{2})/(c_{1)}))/a+(pin)/a