# Thread: help solving trig equation is appreciated

1. ## help solving trig equation is appreciated

solve: 3sin^2 4x = cos 4x - 1

given that 0 deg [greater/equal to] x < 360 deg.

I tried turning it into a quadratic equation using the "1 = sin^2 + cos^2" rule and when that didn't work I pretty much ran out of ideas. Much thanks for any help to be offered.

2. Originally Posted by Savior_Self
solve: 3sin^2 4x = cos 4x - 1

given that 0 deg [greater/equal to] x < 360 deg.

I tried turning it into a quadratic equation using the "1 = sin^2 + cos^2" rule and when that didn't work I pretty much ran out of ideas. Much thanks for any help to be offered.
$\displaystyle 0 \le x < 360$

$\displaystyle 0 \le 4x < 1440$

$\displaystyle 3[1 - \cos^2(4x)] = \cos(4x) - 1$

$\displaystyle 3 - 3\cos^2(4x) = \cos(4x) - 1$

$\displaystyle 0 = 3\cos^2(4x) + \cos(4x) - 4$

$\displaystyle 0 = [3\cos(4x) + 4][\cos(4x) - 1]$

$\displaystyle \cos(4x) = -\frac{4}{3}$ ... no solution

$\displaystyle \cos(4x) = 1$ ...

$\displaystyle 4x = 0$

$\displaystyle 4x = 360$

$\displaystyle 4x = 720$

$\displaystyle 4x = 1080$

you can divide by 4

3. haha, ahh...I'm an idiot. Thanks skeeter.