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

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

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

Move

a c_{2}

to the right side

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

Divide both sides by

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

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

Apply trig inverse properties

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

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

3 - 3 cos (x) = 0

- 2 cot (x) = csc^{2} (x)

sin (θ) = - 0.671

cos (y) = sin (y)

\frac{sin ^{2} ( x ) + 5 cos ^{2} ( x )}{2} = 2

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