# Express $[\cos(x) + \sqrt3 \sin(x)]$ in the form $[r\cos(x-a)]$

Express $[\cos(x) + \sqrt3\sin(x)]$ in the form $[r\cos(x-a)]$, where $r>0$ and $0\leq360$, hence solve the equation $[\cos(x) + \sqrt3\sin(x)= \sqrt2]$

This is as far as i have completed. I don't know whether the question is wrong or i just cant get it

$[\cos(x) + \sqrt3\sin(x)]$

$r=\sqrt{a^2+b^2}$

$r=+-2$

$r>0$

$r=2$

$\tan(a)=\frac {b}{a} =\sqrt3\\a=60^\circ$

therefore $[\cos(x) + \sqrt3\sin(x)]= 2\cos (x-60)$

given that $[\cos(x) + \sqrt3\sin(x)]= 2\cos (x-60)$

$2\cos (x-60)=\sqrt 2$

$\cos (x-60)=(\sqrt2)/2$

$-60 \leq x-60 \leq 300$

I don't know where to go from here... help please

• The angles whos cos is $1/\sqrt{2}$ are $45^\circ$ and relatives. One relative is $-45^\circ$. – André Nicolas Jul 7 '13 at 6:05
• thanks i feel so stupid now – Moneek Jul 7 '13 at 6:05
• Just oversight. I have made a lot more than you. – André Nicolas Jul 7 '13 at 6:09

$\cos x+\sqrt3\sin x=2\cos(x-60^\circ)=\sqrt2$

So, $\cos(x-60^\circ)=\frac1{\sqrt2}=\cos45^\circ$

$\implies x-60^\circ=n360^\circ\pm 45^\circ$ where $n$ is any integer

Taking $'+'$ sign, $x-60^\circ=n360^\circ+45^\circ\implies x=n360^\circ+105^\circ$

If $0\le x\le 360^\circ, 0\le n360^\circ+105^\circ\le 360^\circ\implies n=0$

Taking $'-'$ sign, $x-60^\circ=n360^\circ-45^\circ\implies x=n360^\circ+15^\circ$

If $0\le x\le 360^\circ, 0\le n360^\circ+15^\circ\le 360^\circ\implies n=0$

So, $x=15^\circ,105^\circ$