Question

How do you solve `3sinx+5cosx=4`?

Answer 1

Pose:

`sinx = u`
`cos x = v`

You have:

`3u+5v=4`

and from the fundamental trigonometric identity:

`u^2+v^2=1`

Substitute `v` from the first equation into the second:

`v=(4-3u)/5`

`u^2+((4-3u)/5)^2=1`

`25u^2+16-24u+9u^2-25=0`

`34u^2-24u-9=0`

Using the quadratic formula:

`u=frac (12+-sqrt(144+9*34)) 34=frac (12+-3sqrt(50)) 34`

`u_1=0.97685`
`u_2=-0.27097`

`v_1=0.21389`
`v_2=0.96258`

Both solutions are between `-1` and `1` as the value of a sine or a cosine must be and can be accepted.

The corresponding angles are:

`theta_1=-0.274407416`
`theta_2=1.355246417`

Answer 2

`x=arcsin(4/sqrt(3^2+5^2))-arctan(5/3)`

Explanation:

Making

`3=lambda cos(phi_0)`

and

`5=lambda sin(phi_0)`

with `lambda = sqrt(3^2+5^2)`

we have

`lambda sin(x) cos(phi_0)+lambda cos(x)sin(phi_0)=4`

Now using the identity

`sin(a+b)=sin(a)cos(b)+sin(b)cos(a)` we have

`sin(x+phi_0)=4/lambda=4/sqrt(3^2+5^2)`

so `x+phi_0=arcsin(4/sqrt(3^2+5^2))`

but `phi_0 = arctan(5/3)` then

`x=arcsin(4/sqrt(3^2+5^2))-arctan(5/3)`