How do you solve sin x + 3 cos x = 3?

2 Answers
May 25, 2015

Something you can do is square the whole thing to get some identities going:

(sinx + 3cosx)^2 = 3^2
sin^2x + 6sinxcosx + 9cos^2x = 9
sin^2x + 6sinxcosx = 9-9cos^2x
= 9(1-cos^2x)
= 9(sin^2x)

since sin^2x + cos^2x = 1. Now subtract sin^2x over to get:

8sin^2x = 6sinxcosx

8 = (6cosx)/(sinx) = 6cotx

cotx = 8/6 = 4/3

x = "arccot"(4/3) ~~ 36.87^o ~~ 0.6435 "radians"

Testing it:
sin(0.6435) + 3cos(0.6435) = 3
~0.60 + 3*~0.80 ~~ 0.6 + 2.40 ~~ 3

And it works. Another possible answer is 0:
sin(0) + 3cos(0) = 3
0 + 3*1 = 3

May 25, 2015

Put 3 = tan a --> a = 71.56 deg --> cos a = 0.32

sin x + (sina/cos a).cos x = 3

sin x.cos a. + sin a.cos x = 3cos a = 3(0.32) = 0.95

sin (x + a) = sin (x + 71.56) = 0.95 = sin 71.56

a. x + 71.56 = 71.56 -> x = 0

b. x + 71.56 = 180 - 71.56 = 108.44 -> x = 108.44 - 71.56 = 36.88

Check by equation: sin x + 3cos x = 3.

x = 0 --> 0 + 3(1) = 3. OK

x = 36.88 --> 0.60 + 3(0.80) = 3 OK