Question

What is the general solution for `4cosx+3sinx=-1`?

Answer 1

`x=2(arctan(1/3*(3+2*sqrt(6)))+kpi`

`x=2(arctan(1/3*(3-2sqrt(6)))+kpi`

Explanation:

using the so called Weierstrass Substitution
`sin(x)=(2t)/(1+t^2)`

`cos(x)=(1-t^2)/(1+t^2)`

`tan(x/2)=t`, we get

`4(1-t^2)/(1+t^2)+6t/(1+t^2)=-1` so we get

`0=3t^2-6t-5`

with

`t_{1,2}=1pm2sqrt(2/3)`

Answer 2

`x=2kpi+alpha+-beta,kinZZ,alpha`=#tan^-1 (3/4) , beta#=`cos^-1(-1/5)`

Explanation:

We take,

`4=rcosalpha and 3=rsinalpha...to(1)`

So,

`4cosx+3sinx=-1`

`=>rcosalphacosx+rsinalphasinx=-1`

`=>r[cosxcosalpha+sinxsinalpha]=-1`

`=>rcos(x-alpha)=-1`

`=>color(red)(cos(x-alpha)=-1/r...to(2)`

Form `(1)` ,we get

`color(red)(cosalpha=4/r and sinalpha=3/r...to(3)`

Now,

`cos^2alpha +sin^2alpha=1`

`=>16/r^2+9/r^2=1`

`=>r^2=25=color(blue)(r=5`

From `(3)`

`cosalpha=4/5 > 0 and sinalpha =3/5 > 0`

`=>tanalpha=(3/5)/(4/5)=3/4 >0=>I^(st)Quadrant`

`tanalpha=3/4=>color(orange)(alpha=tan^-1(3/4)`

From `(2)`

`cos(x-alpha)=-1/5`

`cos(x-alpha)=cosbeta ,where,color(violet)( cosbeta=-1/5`

So,the general solution of equn. is :

`x-alpha=2kpi+-beta,kinZZ andbeta=cos^-1(-1/5)`

`:.x=2kpi+alpha+-beta,kinZZ,alpha`=#tan^-1 (3/4) , beta#=`cos^-1(-1/5)`

Answer 3

`x = 138^@41 + k360^@`
`x = 295^@33 + k360^@`

Explanation:

4cos x + 3sin x = -1
Divide both side by 4
`cos x + (3/4)sin x = - 1/4` (1)
Call `tan t = sin t/cos t = 3/4` --> `t = 36^@87` --> `cos t = 0.8`
Equation (1) becomes:
`cos x.cos t + sin t.sin x = - cos t/4`.
`cos (x - t) = cos (x - 36.87) = - 0.8/4 = - 0.2`
Calculator and unit circle give 2 solutions for (x - 36.87)
`(x - 36.87) = +- 101^@54`
a. x - 36.87 = 101.54
`x = 101.54 + 36.87 = 138^@41 + k360^@`
b. x - 36.87 = - 101.54
`x = - 101.54 + 36.87 = - 64^@67 + k360^@`
Note. `x = - 64^@67` is co-terminal to `x = 295^@33`.