Math Help - Solving for X

1. Solving for X

Alright so here is my problem, first day of college started I am in Calculus and have never took a trigonometry class in my life. Because Calculus involves some trigonometry the teacher assigned us a trig review for homework and I have no clue on how to do it, thanks for the help I get.

Here are some of the problems:
Provide all EXACT solutions for the variable on the interval [0,2pie)

2sin3pie/4 = X

2cos(x) = -1

3sec(7pieX/3) = -2(square root of 3)

if you could explain how u got the answers I would appreciate it thank you

2. Originally Posted by Bran21

2sin3pie/4 = X
$X = 2\sin\left(\frac{3\pi}{4}\right)$

$\sin\left(\frac{3\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

Therefore

$X = 2\frac{1}{\sqrt{2}}$

$X = \frac{2}{\sqrt{2}}$

Originally Posted by Bran21

2cos(x) = -1
$2\cos(x) = -1$

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

Cosine is negative in the 2nd and 3rd quadrants so

$x = \pi - \theta, \pi+\theta$ where $\theta = \frac{\pi}{3}$

$x = \pi - \frac{\pi}{3}, \pi+\frac{\pi}{3}$

$x = \frac{2\pi}{3},\frac{4\pi}{3}$

3. Thanks so much pickslides I have a few more questions if you have the time if not youve helped me enough already and im grateful that you did.

Cos(2Pie/3) = 2x

0 = 1 + Sin (3x/2)

Tanx = Cotx

Edit: Also for the second problem how did you come up with theta = pie/3
Also I know its Pi not Pie but Mozilla was fixing it

4. Originally Posted by Bran21

Edit: Also for the second problem how did you come up with theta = pie/3
There are some special triangles you need to commit to memory if you are going to be successful in completing a course in circular trigonometry.

Remember these Special right triangles - Wikipedia, the free encyclopedia

5. Originally Posted by pickslides
There are some special triangles you need to commit to memory if you are going to be successful in completing a course in circular trigonometry.

Remember these Special right triangles - Wikipedia, the free encyclopedia
So looking at this you decided it was 60 degrees and in radians thats pi/3 but I dont understand how you got the 60 degrees then or you looked at the -1/2 and decided from there, im sorry Im really wishing I took a trig course in high school cause its messing me up

6. Assuming you are aware of the different sides on a right angled triangle then.

$\cos(\theta) = \frac{A}{H}$

Consider

$\cos(\theta) = \frac{1}{2}$

and the triangle from the wiki page above. Then the angle that yields $\frac{1}{2}$ for cosine is $\frac{\pi}{3}$

I.e $\cos(\frac{\pi}{3}) = \frac{1}{2}$

Now for

$\cos(\theta) = \frac{-1}{2}$ You have to use the symmetry of the unit circle to find where Cosine is a negative value.

Have a look at

http://www.clarku.edu/~djoyce/trig/functions.html

7. Alright I think im starting to understand this, its a 30-60-90 triangle and that means there is the 60 degree angle and the cos would = 1/2 because the 1 is adjacent on the 30-60-90 and the 2 is the hypotenuse. Then you get the pi/3 from the chart.

Alright now how would I relate tanx = cotx

Thanks for help so far

8. Originally Posted by Bran21
Alright I think im starting to understand this, its a 30-60-90 triangle and that means there is the 60 degree angle and the cos would = 1/2 because the 1 is adjacent on the 30-60-90 and the 2 is the hypotenuse. Then you get the pi/3 from the chart.

Thanks for help so far
Yep

Originally Posted by Bran21

Alright now how would I relate tanx = cotx

Thanks for help so far
Consider $\tan(x) = \frac{\sin(x)}{\cos(x)}$ and $\cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}$

Now $\tan(x) = \cot(x)$ is the same as saying

$\frac{\sin(x)}{\cos(x)} = \frac{\cos(x)}{\sin(x)}$

With a bit of manipulation you get

$\sin^2(x)= \cos^2(x)$

$1-\cos^2(x)= \cos^2(x)$

$1= 2\cos^2(x)$

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

$\cos(x) = \frac{1}{ \sqrt{2}}$

Now go find those triangles again!

9. Originally Posted by pickslides
Yep

Consider $\tan(x) = \frac{\sin(x)}{\cos(x)}$ and $\cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}$

Now $\tan(x) = \cot(x)$ is the same as saying

$\frac{\sin(x)}{\cos(x)} = \frac{\cos(x)}{\sin(x)}$

With a bit of manipulation you get

$\sin^2(x)= \cos^2(x)$

$1-\cos^2(x)= \cos^2(x)$

$1= 2\cos^2(x)$

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

$\cos(x) = {\color{red}\pm}\frac{1}{ \sqrt{2}}$

Now go find those triangles again!
Note my correct in red. You can't neglect the fact that the value can be negative!

10. Thanks Chrissles.

Always rushing to get these things out at work.