Question

cosx -root 3sinx =2

Answer 1

Do you mean this:
\[\cos{x}-\sqrt{3\sin{x}}=2\]

Answer 2

\[\sqrt{3}\sin x\]

Answer 3

Transfer sqrt(3)sin(x) to the other side and square both sides:
\[\cos{x}=2+\sqrt{3}\sin{x}\\
\cos^2{x}=4+4\sqrt{3}\sin{x}+\sin^2{x}\\
1-\sin^2{x}=4+4\sqrt{3}\sin{x}+\sin^2{x}\\
2\ \sin^2{x}+4\sqrt{3}\sin{x}+3=0\]
Now use the quadratic formula.

Answer 4

but don't you get an ugly answer if you do it like that
according to my solution book the answer should be 360n - 60

Answer 5

You wouldn't know until you try.
\[\sin{x}=\frac{-4\sqrt{3}\pm\sqrt{(4\sqrt{3})^2-4(2)(3)}}{2(2)}=\frac{-4\sqrt{3}\pm\sqrt{24}}{4}=-\sqrt{3}\pm\frac{\sqrt{3}}{2}\\
\sin{x}=\frac{-\sqrt{3}}{2},\frac{-3\sqrt{3}}{2}\]
But since the second one is out of the range of sin(x), we're left with \[\sin{x}=\frac{-\sqrt{3}}{2}\]
See? It wasn't so bad. :D

Answer 6

\[\cos x - \sqrt{3} \sin x = 2\]\[\cos x - \tan 60^{\circ} \sin x = 2\]multiply both sides with cos 60°
\[\cos 60^{\circ} \cos x - \sin 60^{\circ} \sin x = 1\]\[\cos \left( 60^{\circ} + x \right) = \cos 0^{\circ}\]\[60^{\circ} + x = n 360^{\circ}\]\[x = n360^{\circ} - 60^{\circ}\]

Answer 7

thank you to both ! help greatly appreciated! =)