I'll use x in place of θ
Let's isolate the sine term and then square both sides
3sin(x) + 4cos(x) = 5
3sin(x) = 5-4cos(x)
(3sin(x))^2 = ( 5-4cos(x) )^2
9sin^2(x) = 25-40cos(x)+16cos^2(x)
9sin^2(x) = 25-40cos(x)+16(1-sin^2(x))
9sin^2(x) = 25-40cos(x)+16-16sin^2(x)
9sin^2(x)+16sin^2(x) = 25-40cos(x)+16
25sin^2(x) = 41-40cos(x)
25(1-cos^2(x)) = 41-40cos(x)
25-25cos^2(x) = 41-40cos(x)
Now let w = cos(x), so we now have the following
25-25w^2 = 41-40w
25w^2-40w+41-25 = 0
25w^2-40w+16 = 0
(5w - 4)^2 = 0
5w-4 = 0
w = 4/5
Therefore cos(x) = 4/5
Use the pythagorean theorem or unit circle to find that if cos(x) = 4/5, then it leads to sin(x) = 3/5. This is exactly because of the 3-4-5 right triangle.
Then lastly,
tan(x) = sin(x)/cos(x)
tan(x) = sin(x) ÷ cos(x)
tan(x) = (3/5) ÷ (4/5)
tan(x) = (3/5) * (5/4)
tan(x) = (3*5)/(5*4)
tan(x) = 3/4