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

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Answer 1 · 2015-05-25T22:45:59

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#

Answer 2 · 2015-05-25T22:56:10

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