Question

How to find the general solution 5 sin(x)+2 cos(x)=3?

Answer 1

`rarrx=npi+(-1)^n*(sin^(-1)(3/sqrt29))-sin^(-1)(2/sqrt29)` `n inZZ`

Explanation:

`rarr5sinx+2cosx=3`

`rarr(5sinx+2cosx)/(sqrt(5^2+2^2))=3/(sqrt(5^2+2^2)`

`rarrsinx*(5/sqrt(29))+cosx*(2/sqrt(29))=3/sqrt29`

Let `cosalpha=5/sqrt29` then `sinalpha=sqrt(1-cos^2alpha)=sqrt(1-(5/sqrt29)^2)=2/sqrt29`

Also, `alpha=cos^(-1)(5/sqrt29)=sin^(-1)(2/sqrt29)`

Now, given equation transforms to

`rarrsinx*cosalpha+cosx*sinalpha=3/sqrt29`

`rarrsin(x+alpha)=sin(sin^(-1)(3/sqrt29))`

`rarrx+sin^(-1)(2/sqrt29)=npi+(-1)^n*(sin^(-1)(3/sqrt29))`

`rarrx=npi+(-1)^n*(sin^(-1)(3/sqrt29))-sin^(-1)(2/sqrt29)` `n inZZ`

Answer 2

`x = 12^@12 + k360^@`
`x = 124^@28 + k360^@`

Explanation:

5sin x + 2cos x = 3.
Divide both sides by 5.
`sin x + 2/5 cos x = 3/5 = 0.6` (1)
Call `tan t = sin t/(cos t) = 2/5` --> `t = 21^@80` --> cos t = 0.93.
The equation (1) becomes:
`sin x.cos t + sin t.cos x = 0.6(0.93)`
`sin (x + t) = sin (x + 21.80) = 0.56`
Calculator and unit circle give 2 solutions for (x + t) -->
a. x + 21.80 = 33.92
`x = 33.92 - 21.80 = 12^@12`
b. x + 21.80 = 180 - 33.92 = 146.08
`x = 146.08 - 21.80 = 124^@28`
General answers:
`x = 12^@12 + k360^@`
`x = 124^@28 + k360^@`
Check by calculator.
`x = 12^@12` --> 5sin x = 1.05 --> 2cos x = 1.95
5sin x + 2cos x = 1.05 + 1.95 = 3. Proved.
`x = 124^@28` --> 5sin x = 4.13 --> 2cos x = -1.13
5sin x + 2cos x = 4.13 - 1.13 = 3. Proved.