3
(a) I do not know the proof, but there is a trig identity that says that
a sin x + b cos x = c sin (x + α) where c = sqrt (a^2 + b^2) and
α = atan2(b,a)
So
c = sqrt (4^2 + 3^2) = sqrt (25) = 5 and
atan2 (b , a) = atan2( 3 , 4) ≈ .6435 (in rads)
4 sin x + 3 cos x = 5 sin ( x + .6435 )
(b) 4sin x + 3 cos x = 2
4sin x = 2 - 3 cos x square both sides
16sin^2 x = 4 - 12cosx + 9cos^2 x
16(1 - cos^2 x) = 4 - 12cos x + 9cos^2 x
16 - 16 cos^2 x = 4 - 12cos x + 9cos^2 x
25 cos^2x - 12 cos x - 12 = 0 let cos x = a
25a^2 - 12 a - 12 = 0
Using the quad formula.......
a = [ 6 ± 4 sqrt (21)] / 25
So
a = cos x = [ 6 ± 4 sqrt (21)] / 25
Wnen cos x = [ 6 + 4 sqrt (21)] / 25
arccos [ [ 6 + 4 sqrt (21)] / 25 ] = x ≈ .232 rads and ≈ 6.051 rads
When cos x = arccos [ [ 6 - 4 sqrt (21)] / 25 ]
arccos [ [ 6 - 4 sqrt (21)] / 25 ] = x ≈ 2.0866 rads and ≈ 5.228 rads
However.......232 rads and 5.228 rads are extraneous [ because of squaring ]
As the graph here shows.....the two solutions are x ≈ 2.0866 rads and x ≈ 6.051 rads
https://www.desmos.com/calculator/pjvewx15zw
