What is the general solution for sin(A)+3cos(A)=0 ?

The general solution for sin(A)+3cos(A)=0 is A=-1.24904…+pin

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sin(A)+3cos(A)=0

sin (A) + 3 cos (A) = 0

A = - 1.24904 \dots + πn

Degrees

A = - 71.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

sin (A) + 3 cos (A) = 0

Rewrite using trig identities

sin (A) + 3 cos (A) = 0

Divide both sides by

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

\frac{sin ( A ) + 3 cos ( A )}{cos ( A )} = \frac{0}{cos ( A )}

Simplify

\frac{sin ( A )}{cos ( A )} + 3 = 0

Use the basic trigonometric identity:

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

tan (A) + 3 = 0

tan (A) + 3 = 0

Move

3

to the right side

tan (A) + 3 = 0

Subtract

3

from both sides

tan (A) + 3 - 3 = 0 - 3

Simplify

tan (A) = - 3

tan (A) = - 3

Apply trig inverse properties

tan (A) = - 3

General solutions for

tan (A) = - 3

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

A = arctan (- 3) + πn

A = arctan (- 3) + πn

Show solutions in decimal form

A = - 1.24904 \dots + πn

x, 3 cot (x) - 1 = 0

- sin (x) = 3 cos (x)

cos (2 θ) = sin (θ), 0 \leq θ < 2 π

sin (x) (1 + cos (x)) = sin (x) + cos (x)

9 0^{2} = 6 0^{2} + 7 0^{2} - 2 (60) (70) (cos (x))

What is the general solution for sin(A)+3cos(A)=0 ?
The general solution for sin(A)+3cos(A)=0 is A=-1.24904…+pin