4sinΘ - 3cosΘ = 2?

Question

Solve.......4sinΘ - 3cosΘ = 2

Answer 1

Let 4 = x cos r
3 = x sin r
then x = 5
and sin r =3/5

now 4 sin t - 3 cos t t for theta
= 5 cos r sin t - 5 sin r cos t
= 5 sin(t-r) = 2

t-r = sin ^-1(2/5)
so t = sin^-1 (2/5) - sin ^-1(3/5)

Answer 2

I'll drop the theta notation for x, then,

4sin x - 3cos x =2
=> 5 [(4/5)sin x - (3/5)cos x] = 2 , as the expression is unchanged.
Now let φ = arccos (4/5) = arcsin (3/5) , then,

5.[cos φ . sin x - sin φ . cos x] =2
that is sin (x - φ) = 2/5
x = φ + n.pi + [(-1)^n].arcsin (2/5)
where n is an integer.
Note that I've taken the principle solution for φ.
when n=0 you get the same for x, namely

x = φ + arcsin (2/5) = arcsin (3/5) + arcsin (2/5)

Answer 3

4sinx-3cosx=2
this is of the type acosx+bsinx=c where abs.value of c<(a^2+b^2)^1/2.here a=-3,b=4 and c=2
dividing the equation by 5
-3/5cos+4/5sinx=2/5
letfor least angle y cosy=-3/5 and siny=4/5
so substituting
cosycox+sinysinx=2/5
cos(y-x)=2/5
let for the smallest angle p cosp2/5
cos(y-x)=cosp
therefore y-x=2npi+/-p
or y=2npi+/-p+x