Find the exact value? #3-2sin^2x=3cosx#

2 Answers

#rarrx=2npi+-pi/3# OR #x=2npi#

Explanation:

#rarr3-2sin^2x=3cosx#

#rarr3-2(1-cos^2x)=3cosx#

#rarr3-2+2cos^2x=3cosx#

#rarr2cos^2x-3cosx+1=0#

#rarr2cos^2x-2cosx-cosx+1=0#

#rarr2cosx(cosx-1)-1(cosx-1)=0#

#rarr(2cosx-1)(cosx-1)=0#

Either, #2cosx-1=0#

#rarrcosx=1/2=cos(pi/3)#

#rarrx=2npi+-pi/3# where #nrarrZ#

OR, #cosx-1=0#

#rarrcosx=1=cos 0#

#rarrx=2pi n# where #nrarrZ#

Jun 2, 2018

#x = pm 60^circ + 360^circ k or x = 360^circ k quad# integer #k #

Explanation:

The other answer is fine; I just want to add this to my collection of yet anther trig problem that's going to work out to 30/60/90 or 45/45/90. When are we going to realize that it's goofy to teach a subject that only handles two triangles reasonably well?

#3 - 2 sin ^2 x = 3 cos x#

#3 - 2(1 - cos^2 x) =3 cos x#

#2 cos ^2x - 3 cos x + 1 = 0#

#(2 cos x - 1 )(cos x - 1) = 0#

#cos x = 1/2 or cos x = 1#

Cosine or sine of #1/2#; same old same old.

#cos x = cos 60^circ or cos x = cos 0 #

#x = pm 60^circ + 360^circ k or x = 360^circ k quad# integer #k #