What is the general solution for asin(x)+bcos(x)=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

asin(x)+bcos(x)=0

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

x = arctan (- \frac{b}{a}) + πn

Solution steps

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

Rewrite using trig identities

a tan (x) + b = 0

Move

b

to the right side

a tan (x) = - b

Divide both sides by

a; a \neq = 0

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

Apply trig inverse properties

x = arctan (- \frac{b}{a}) + πn

sin (2 θ) = \frac{4}{5}

tan (2 x) = - 0.025

4 7^{\circ} = cos (x), x

cos (α) = \frac{7}{11}

x, m^{2} - m - 2 = sin (x)

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