What is the general solution for acos(x)+bsin(x)=0 ?

The general solution for acos(x)+bsin(x)=0 is x=arctan(-a/b)+pin

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acos(x)+bsin(x)=0

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

x = arctan (- \frac{a}{b}) + πn

Solution steps

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

Rewrite using trig identities

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

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

a + b tan (x) = 0

a + b tan (x) = 0

Move

a

to the right side

a + b tan (x) = 0

Subtract

a

from both sides

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

Simplify

b tan (x) = - a

b tan (x) = - a

Divide both sides by

b; b \neq = 0

b tan (x) = - a

Divide both sides by

b; b \neq = 0

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

Simplify

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

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

Apply trig inverse properties

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

General solutions for

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

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

x = arctan (- \frac{a}{b}) + πn

x = arctan (- \frac{a}{b}) + πn

cot (x) = \frac{7}{4}, sin (x) < 0

22 cos (x) + 2 = 0

8 cos (2 x) = 1

cos (- (\frac{π}{6}) - 2 x) = 1

csc (x) = 6

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