How do you solve #2cosx+2sinx=sqrt6# in the interval [0,360]?

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Answer 1 · 2016-09-01T18:38:07

The solutions in the given interval are #x = 15˚# and #x = 75˚#.

Factor out a 2:

#2(cosx + sinx) = sqrt(6)#

#cosx + sinx = sqrt(6)/2#

Square both sides:

#(cosx + sinx)^2 = (sqrt(6)/2)^2#

#cos^2x + 2sinxcosx + sin^2x = 6/4#

Apply the identity #sin^2x + cos^2x = 1#:

#1 + 2sinxcosx = 6/4#

Apply the identity #2sinxcosx = sin2x#:

#sin2x = 1/2#

#2x = 30˚ and 2x = 150˚#

#x = 15˚ and 75˚#

Check both solutions in the original equation. Both work. Hence, our solution set is #{15˚, 75˚}#.

Hopefully this helps!