What is the general solution for a*sin(x)+b*cos(x)=0 ?

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

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asin(x)+bcos(x)=0

a \cdot sin (x) + b \cdot cos (x) = 0

x = arctan (- \frac{b}{a}) + πn

Solution steps

a sin (x) + b cos (x) = 0

Rewrite using trig identities

a sin (x) + b cos (x) = 0

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

a tan (x) + b = 0

a tan (x) + b = 0

Move

b

to the right side

a tan (x) + b = 0

Subtract

b

from both sides

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

Simplify

a tan (x) = - b

a tan (x) = - b

Divide both sides by

a; a \neq = 0

a tan (x) = - b

Divide both sides by

a; a \neq = 0

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

Simplify

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

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

Apply trig inverse properties

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

General solutions for

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

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

x = arctan (- \frac{b}{a}) + πn

x = arctan (- \frac{b}{a}) + πn

cos (θ) = - 0.19

9 cos^{2} (x) = 2

tan (θ) = \frac{6}{14}

tan (θ) = \frac{6}{15}

tan (x) = \frac{10}{3}

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