Question

How do you solve `3sin theta - 4 cos theta = 2`?

Answer 1

`S={209.551...+360^@n, 76.708...+360^@n}`

Explanation:

Use the following formulas to transform the equation then solve
`A cos x + B sin x = C cos (x – D)`

`C = sqrt(A^2+B^2 ), cos D = A/C and sin D = B/C`

`A=-4, B=3,C=sqrt(16+9)=sqrt25=5`

`cosD=-4/5->D=cos^-1 (-4/5) =143.13...^@`-> Quadrant II

`5cos(theta-143.13...^@)=2`

`cos(theta-143.13...^@)=2/5`

#theta-143.13...^@=cos^-1(2/5)#

`theta-143.13...^@=+-66.4218...+360^@n`

`theta=143.13...+-66.4218...+360^@n`

`theta = 209.551...+360^@n, 76.708...+360^@n`

`S={209.551...+360^@n, 76.708...+360^@n}`

Answer 2

`x = 76^@71 + 2kpi`
`x = 209.55 + 2kpi`

Explanation:

Divide both sides by 3 -->
`sin x - (4/3)cos x = 2/3`
Call `tan u = (sin u)/(cos u) = 4/3`--> `u = 53^@13`
`sin x.cos u - sin u.cos x = (2/3)cos u = (2/3)0.60 = 0.40`
Apply the trig identity: sin (a - b) = sin a.cos b - sin b.cos a -->
sin (x - 53.13) = 0.40 --> sin (23.58) and sin (180 - 23.58)
Calculator and unit circle give 2 solution arcs:

`a. (x - 53.13) = 23^@58` --> `x = 23.58 + 53.13 = 76^@71`
`b. (x - 53.13) = 180 - 23.58 = 156^@42`--> x = 156.42 + 53.13 =
`= 209^@55`