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 tan (θ) = b

Divide both sides by

a; a \neq = 0

tan (θ) = \frac{b}{a}; a \neq = 0

Apply trig inverse properties

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