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
$\sqrt{3} \cdot \sin{x} = 2 - \cos{x}$

square both sides ...

$3\sin^2{x} = 4 - 4\cos{x} + \cos^2{x}$

use the Pythagorean identity ...

$3(1 - \cos^2{x}) = 4 - 4\cos{x} + \cos^2{x}$

$3 - 3\cos^2{x} = 4 - 4\cos{x} + \cos^2{x}$

$0 = 4\cos^2{x} - 4\cos{x} + 1
$

$0 = (2\cos{x} - 1)^2$

$\cos{x} = \frac{1}{2}$

$x = \frac{\pi}{3} \, , \, \frac{5\pi}{3}$

check for extraneous solution(s) (see if they work in the original equation)

3. Hello, shane99!

Another approach . . . clever, but obscure.

Solve for $x\!:\;\;\sqrt{3}\sin x + \cos x \:=\:2$

Divide by 2: . $\tfrac{\sqrt{3}}{2}\sin x + \tfrac{1}{2}\cos x \;=\;1$

Note that: . $\begin{array}{ccc}\sin\frac{\pi}{6} &=& \frac{1}{2} \\ \\[-4mm] \cos\frac{\pi}{6} &=& \frac{\sqrt{3}}{2} \end{array}$

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

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

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

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

4. Originally Posted by Soroban

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

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

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

which equals

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