1. Trigonometric identities

√3 sinx+cosx=2, solve for x, please explain how I have to do this

2. Originally Posted by shane99
√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)

3. 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: .$\displaystyle \underbrace{\cos\tfrac{\pi}{6}\sin x + \sin\tfrac{\pi}{6}\cos x} \:=\:1$

. . . . . . . . . . . . . $\displaystyle \sin(x + \tfrac{\pi}{6}) \:=\:1$

. . . . . . . . . . . . . . . .$\displaystyle x + \tfrac{\pi}{6} \:=\:\tfrac{\pi}{2}$

. . . . . . . . . . . . . . . . . . $\displaystyle x \:=\:\frac{\pi}{3}$

4. Originally Posted by Soroban

So we have: .$\displaystyle \underbrace{\cos\tfrac{\pi}{6}\sin x + \sin\tfrac{\pi}{6}\cos x} \:=\:1$

. . . . . . . . . . . . . $\displaystyle \sin(x + \tfrac{\pi}{6}) \:=\:1$
What I am confused about is that after you plugged in those identities into the equation how did you end up with sin(x+π/6) = 1. Can you explain what you did to get that equation, because where did cos go.

5. Originally Posted by shane99
What I am confused about is that after you plugged in those identities into the equation how did you end up with sin(x+π/6) = 1. Can you explain what you did to get that equation, because where did cos go.
Take the sine inverse of both sides and you get

$\displaystyle x + \frac{\pi}{6} = \sin^{-1}{(1)}$

which equals

$\displaystyle x + \frac{\pi}{6} = \frac{\pi}{4}$