1. ## trigonometry

solve for x
2 Sin^2X + 2 SinX CosX - 1= 0

2. Dear william,

use the following identity:

$2sin(x)cos(x) = sin(2x)$
$1 = cos^2(x) + sin^2(x)$
$cos^2(x) - sin^2(x) = cos(2x)$

3. Hello, william!

You should be familiar with these identities:

. . $\sin^2\theta \:=\:\frac{1-\cos2\theta}{2} \quad\Rightarrow\quad 2\sin^2\!\theta \:=\:1 - \cos2\theta$

. . $2\sin\theta\cos\theta \:=\:\sin2\theta$

Solve for $x\!:\;\;2\sin^2\!x + 2\sin x\cos x - 1\:=\: 0$

We have: . $\underbrace{2\sin^2\!x} + \underbrace{2\sin x\cos x} -1 \;=\;0$

. . . . . . . $\overbrace{1-\cos2x} + \overbrace{\sin2x} - 1 \;=\;0$

. . $\sin2x \:=\:\cos 2x \quad\Rightarrow\quad \frac{\sin2x}{\cos2x} \:=\:1 \quad\Rightarrow\quad \tan2x \:=\:1$

Therefore: . $2x \:=\:\frac{\pi}{4} + \pi n \quad\Rightarrow\quad x \;=\;\frac{\pi}{8} + \frac{\pi}{2}n\;\;\text{ for any integer }n$

4. Originally Posted by william
solve for x
2 Sin^2X + 2 SinX CosX - 1= 0
sin 2x = 2 sin x cos x
and
cos 2x = cos2x - sin2x = 1 - 2 sin2x
so
2 sin2x + 2 sin x cos x -1 = 0
is equivalent to
sin 2x - cos 2x = 0
which is
sin 2x = cos 2x
or
tan 2x = 1.
therefore, in the first quadrant, 2x is 45o