√3 sinx+cosx=2, solve for x, please explain how I have to do this
$\displaystyle \sqrt{3} \cdot \sin{x} = 2 - \cos{x}$
square both sides ...
$\displaystyle 3\sin^2{x} = 4 - 4\cos{x} + \cos^2{x}$
use the Pythagorean identity ...
$\displaystyle 3(1 - \cos^2{x}) = 4 - 4\cos{x} + \cos^2{x}$
$\displaystyle 3 - 3\cos^2{x} = 4 - 4\cos{x} + \cos^2{x}$
$\displaystyle 0 = 4\cos^2{x} - 4\cos{x} + 1
$
$\displaystyle 0 = (2\cos{x} - 1)^2$
$\displaystyle \cos{x} = \frac{1}{2}$
$\displaystyle x = \frac{\pi}{3} \, , \, \frac{5\pi}{3}$
check for extraneous solution(s) (see if they work in the original equation)
Hello, shane99!
Another approach . . . clever, but obscure.
Solve for $\displaystyle x\!:\;\;\sqrt{3}\sin x + \cos x \:=\:2$
Divide by 2: .$\displaystyle \tfrac{\sqrt{3}}{2}\sin x + \tfrac{1}{2}\cos x \;=\;1$
Note that: .$\displaystyle \begin{array}{ccc}\sin\frac{\pi}{6} &=& \frac{1}{2} \\ \\[-4mm] \cos\frac{\pi}{6} &=& \frac{\sqrt{3}}{2} \end{array}$
So we have: .$\displaystyle \underbrace{\cos\tfrac{\pi}{6}\sin x + \sin\tfrac{\pi}{6}\cos x} \:=\:1$
. . . . . . . . . . . . . $\displaystyle \sin(x + \tfrac{\pi}{6}) \:=\:1$
. . . . . . . . . . . . . . . .$\displaystyle x + \tfrac{\pi}{6} \:=\:\tfrac{\pi}{2}$
. . . . . . . . . . . . . . . . . . $\displaystyle x \:=\:\frac{\pi}{3}$