How do you solve # -sqrt(3)cos x + sin x = -sqrt(3) #?

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Answer 1 · 2018-07-09T21:30:29

#x=2n\pi\pm\pi/6-\pi/6#

Given that

#-\sqrt3\cos x+\sin x=-\sqrt3#

#\sqrt3\cos x-\sin x=\sqrt3#

#(\sqrt3/2)\cos x-(1/2)\sin x=\sqrt3/2#

#\cos x\cos(\pi/6)-\sin x\sin(\pi/6)=\cos(\pi/6)#

#\cos (x+\pi/6)=\cos(\pi/6)#

#(x+\pi/6)=2n\pi\pm\pi/6#

#x=2n\pi\pm\pi/6-\pi/6#

Where, #n# is any integer i.e. #n=0, \pm1, \pm2, \pm3, \ldots#

Answer 2 · 2018-07-09T21:50:22

#x = 2 pi n " and "x = 2 pi n + (5 pi)/3#

Given: #-sqrt(3) cos x + sin x = - sqrt(3)#

Rearrange: #" "sqrt(3) - sqrt(3) cos x + sin x = 0#

When we think of #sqrt(3)# in terms of cosine, realize #cos(pi/6) = sqrt(3)/2#

Group the second two terms and multiply and divide both terms by #2/2#:

#sqrt(3) - 2((sqrt(3))/2 cos x - 1/2sin x) = 0#

Substitute in #cos (pi/6) " for " sqrt(3)/2 " and " sin (pi/6) " for " 1/2:#

#sqrt(3) - 2(cos (pi/6) cos x - sin (pi/6) * sin x) = 0#

Realize that #" "cos(x + pi/6) = cos (pi/6) cos x - sin (pi/6) * sin x#

#sqrt(3) - 2cos(x + pi/6) = 0#

#- 2cos(x + pi/6) = -sqrt(3)#

#cos(x + pi/6) = sqrt(3)/2#

Take the inverse cosine of both sides:

#cos^-1(cos(x + pi/6)) = cos^-1(sqrt(3)/2)#

#x + pi/6 = 2 pi n + pi/6#

and #x + pi/6 = 2 pi n + (11 pi)/6#

Simplify:

#x = 2 pi n " and " x = 2 pi n +(10 pi/6) = 2 pi n + (5 pi)/3#