Question

cosx -root3sinx =2

Answer 1

\[cosx -\sqrt3sinx =2\]\[cosx =2+\sqrt3sinx \]Square both sides\[cos^2x =(2+\sqrt3sinx)^2 \]\[cos^2x =4+4\sqrt3sinx+ 3sin^2x \]\[1-sin^2x =4+4\sqrt3sinx+ 3sin^2x \]\[0 =3+4\sqrt3sinx+ 4sin^2x \]Time to use quadratic formula...
\[sinx = (-4\sqrt3 \pm \sqrt{(4\sqrt3)^2-4(4)(3)})/2(4)\]\[sinx = (-4\sqrt3)/8\]\[sinx = (-\sqrt3)/2\]
x = 180 +60 or x=360-60
x=240 (rejected) or x=300

Not sure if it is correct :S

Answer 2

THANKS BIG TIME!

Answer 3

welcome, but you'd better check it :S

Answer 4

YOU ARE CORRECT ACCORDING TO MY SOLUTION BOOKLET! =D

Answer 5

hope that the solution in your book is correct too ... :)

Answer 6

without quadratic formula\[\cos x-\sqrt{3}\sin x=2\]\[\cos x+\tan \frac{2\pi}{3}\sin x=2\]\[\cos \frac{2\pi}{3}\cos x+\sin \frac{2\pi}{3}\sin x=2\cos \frac{2\pi}{3}\]\[\cos \left( \frac{2\pi}{3}-x \right)=-1\]\[\cos \left( \frac{2\pi}{3}-x \right)=\cos \pi=\cos \left( -\pi \right)\]\[x = -\frac{\pi}{3}=-60^{\circ}\]or\[x = \frac{5\pi}{3}=300^{\circ}\]reject the negative x

Answer 7

actually, -60 = 300 (both in degrees)...

Answer 8

true, because\[\cos \left( 360^{\circ}-x \right)=\cos x=\cos \left( -x \right)\]

Answer 9

divide both sides of the equation by 2
0.5cosx - sqrt(3)/2 sin x =1
cos60.cosx-sin60.sinx=1
cos(60+x)=1=cos0
60+x=0 or 60+x=360
x=-60 x=300 ,see if you understand????