3sintheta + 4costheta = 0
Let 3sintheta + 4costheta = Rcos(theta-alpha)
= Rcostheta.cosalpha+Rsintheta.sinalpha
By comparison, 3sintheta = Rsintheta.sinalpha
3 = Rsinalpha
By comparison, 4costheta = Rcostheta.cosalpha
4 = Rcosalpha
Therefore:
Rsinalpha = 3------Eq1 (Quadrants 1,2)
Rcosalpha = 4------Eq2 (Quadants 1,4)
Hence, Quadrant 1
(Eq1)^2 + (Eq2)^2=
R^2(sin^2alpha+cos^2alpha)= 3^2 + 4^2
Since sin^2alpha+cos^2alpha = 1,
R^2 = 25
R=5 (R>0)
Rsinalpha = 3------Eq1 (Quadrants 1,2)
Rcosalpha = 4------Eq2 (Quadants 1,4)
Eq1 divided by Eq2
tanalpha = 3/4
alpha = tan^-1(3/4) (Take the acute angle, since quadrant 1)
Therefore,
3sintheta + 4costheta = 5cos(theta - tan^-1(3/4))
Since 3sintheta + 4costheta = 0, 5cos(theta - tan^-1(3/4)) = 0
Solve for 5cos(theta - tan^-1(3/4)) =0.
5cos(theta - tan^-1(3/4)) = 0
cos(theta - tan^-1(3/4)) = 0
theta = 180n + tan^-1(3/4), where n is an integer
Then, substitue values for n for angles between your questions given condition
Hope this helped