What is the general solution for solvefor a,cos(ax)+(b/x)sin(ax)=0 ?

The general solution for solvefor a,cos(ax)+(b/x)sin(ax)=0 is a=(arctan(-x/b))/x+(pin)/x

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solvefor a,cos(ax)+(b/x)sin(ax)=0

solve for

a, cos (a x) + (\frac{b}{x}) sin (a x) = 0

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

Solution steps

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

Rewrite using trig identities

1 + \frac{b tan ( a x )}{x} = 0

Move

1

to the right side

\frac{b tan ( a x )}{x} = - 1

Multiply both sides by

x

b tan (a x) = - x; x \neq = 0

Divide both sides by

b; x \neq = 0, b \neq = 0

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

Apply trig inverse properties

a x = arctan (- \frac{x}{b}) + πn

Solve

a x = arctan (- \frac{x}{b}) + πn : a = \frac{arctan ( - \frac{x}{b} )}{x} + \frac{πn}{x}; x \neq = 0

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

x, csc (2 x) = \frac{1}{2} (sec (θ)) (cos (θ))

sin (θ) = \frac{3 2 - 3}{40}

cos^{2} (x) + 5 cos (x) - 3 = 0

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

cos^{4} (a) + cos^{2} (a) + sin^{2} (a) + sin^{2} (a) = 1

What is the general solution for solvefor a,cos(ax)+(b/x)sin(ax)=0 ?
The general solution for solvefor a,cos(ax)+(b/x)sin(ax)=0 is a=(arctan(-x/b))/x+(pin)/x