What is the general solution for asin(θ)=bcos(θ) ?

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

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asin(θ)=bcos(θ)

a sin (θ) = b cos (θ)

θ = arctan (\frac{b}{a}) + πn

Solution steps

a sin (θ) = b cos (θ)

Rewrite using trig identities

a sin (θ) = b cos (θ)

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{a sin ( θ )}{cos ( θ )} = \frac{b cos ( θ )}{cos ( θ )}

Simplify

\frac{a sin ( θ )}{cos ( θ )} = b

Use the basic trigonometric identity:

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

a tan (θ) = b

a tan (θ) = b

Divide both sides by

a; a \neq = 0

a tan (θ) = b

Divide both sides by

a; a \neq = 0

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

Simplify

tan (θ) = \frac{b}{a}; a \neq = 0

tan (θ) = \frac{b}{a}; a \neq = 0

Apply trig inverse properties

tan (θ) = \frac{b}{a}

General solutions for

tan (θ) = \frac{b}{a}

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

θ = arctan (\frac{b}{a}) + πn

θ = arctan (\frac{b}{a}) + πn

- 9 tan^{2} (θ) + 12 tan (θ) - 4 = 0

- 5 cos (x) = 0

2 sin^{2} (θ) - sin (θ) = 3

tan (a) = \frac{5}{4}

4 sin (2 x) - 3 = 0

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