# Thread: Help solving trigometric equations!

1. ## Help solving trigometric equations!

1. 4cos^2 3x = 1
2. cos(2x-pi/2)=root2/2
3. cos2x=sinx+1
4. sin6xcosx-cos6xsinx= - root3/2
4. cos2x-4=3cosx-4sin^2x

Please explain to me how to solve these!!

Also, how can i type the root or squares out like I see people do in this forum? or is it an image?

2. Originally Posted by dorkymichelle

Also, how can i type the root or squares out like I see people do in this forum? or is it an image?
http://www.mathhelpforum.com/math-he...-tutorial.html

3. Originally Posted by dorkymichelle
1. 4cos^2 3x = 1
2. cos(2x-pi/2)=root2/2
3. cos2x=sinx+1
4. sin6xcosx-cos6xsinx= - root3/2
4. cos2x-4=3cosx-4sin^2x

Please explain to me how to solve these!!

Also, how can i type the root or squares out like I see people do in this forum? or is it an image?
For the mathematical typesetting, it's using an inbuilt LaTeX compiler. Click on an image and its code comes up.

1. $\displaystyle 4\cos^2{(3x)} = 1$

$\displaystyle \cos^2{(3x)} = \frac{1}{4}$

$\displaystyle \cos{(3x)} = \pm \frac{1}{2}$

$\displaystyle 3x = \arccos{\left(\pm \frac{1}{2}\right)}$

$\displaystyle 3x = \left\{\frac{\pi}{3}, \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3}, 2\pi - \frac{\pi}{3}\right\} + 2\pi n, n \in \mathbf{Z}$

$\displaystyle 3x = \left \{ \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\right\} + 2\pi n$

$\displaystyle x = \left \{ \frac{\pi}{9}, \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{5\pi}{9}\right\} + \frac{2\pi n}{3}$.

4. Originally Posted by dorkymichelle
1. 4cos^2 3x = 1
2. cos(2x-pi/2)=root2/2
3. cos2x=sinx+1
4. sin6xcosx-cos6xsinx= - root3/2
4. cos2x-4=3cosx-4sin^2x

Please explain to me how to solve these!!

Also, how can i type the root or squares out like I see people do in this forum? or is it an image?
2. $\displaystyle \cos{\left(2x - \frac{\pi}{2}\right)} = \frac{\sqrt{2}}{2}$

$\displaystyle 2x - \frac{\pi}{2} = \arccos{\left(\frac{\sqrt{2}}{2}\right)}$

$\displaystyle 2x - \frac{\pi}{2} = \left\{ \frac{\pi}{6}, 2\pi - \frac{\pi}{6}\right\} + 2\pi n, n \in \mathbf{Z}$

$\displaystyle 2x - \frac{\pi}{2} = \left \{ \frac{\pi}{6}, \frac{11\pi}{6} \right\} + 2\pi n$

$\displaystyle 2x = \left \{ \frac{2\pi}{3}, \frac{7\pi}{3}\right\} + 2\pi n$

$\displaystyle x = \left \{ \frac{\pi}{3}, \frac{7\pi}{6} \right\} + \pi n$.

5. Originally Posted by dorkymichelle
1. 4cos^2 3x = 1
2. cos(2x-pi/2)=root2/2
3. cos2x=sinx+1
4. sin6xcosx-cos6xsinx= - root3/2
4. cos2x-4=3cosx-4sin^2x

Please explain to me how to solve these!!

Also, how can i type the root or squares out like I see people do in this forum? or is it an image?
3. $\displaystyle \cos{(2x)} = \sin{x} + 1$

$\displaystyle 1 - 2\sin^2{x} = \sin{x} + 1$

$\displaystyle 0 = 2\sin^2{x} + \sin{x}$

$\displaystyle 0 = \sin{x}(2\sin{x} + 1)$

Therefore

$\displaystyle \sin{x} = 0$ or $\displaystyle 2\sin{x} + 1 = 0$

Case 1:

$\displaystyle \sin{x} = 0$

$\displaystyle x = \arcsin{0}$

$\displaystyle x = \left \{ 0, \pi \right\} + 2\pi n, n \in \mathbf{Z}$.

Case 2:

$\displaystyle 2\sin{x} + 1 = 0$

$\displaystyle 2\sin{x} = -1$

$\displaystyle \sin{x} = -\frac{1}{2}$

$\displaystyle x = \arcsin{\left(-\frac{1}{2}\right)}$

$\displaystyle x = \left \{ \pi - \frac{\pi}{6}, \pi + \frac{\pi}{6} \right \} + 2\pi n, n \in \mathbf{Z}$

$\displaystyle x = \left \{ \frac{5\pi}{6}, \frac{7\pi}{6} \right \} + 2\pi n$.

Putting them together we get

$\displaystyle x = \left \{0, \frac{5\pi}{6}, \pi, \frac{7\pi}{6} \right \} + 2\pi n$.

6. Okay, I'll repost these with my attempts, which to warn you guys.. are horrible!