1. ## Trigonometry

Given that $\displaystyle cos 3x = 4cos^3x -3cosx$,

(a) Find the exact value of $\displaystyle 8cos^310 - 6cos10$
(b) solve the equation $\displaystyle 8cos^3x - 6cosx = - 1$ for 0< x < 180

Attempted solution

(a) 8cos^310 - 6cos10 = 2(cos3(10))
= 2cos30
= $\displaystyle \sqrt{3}$

(b) since 8cos^3x - 6cosx = 2cos 3x
2cos3x = -1
I cosine inversed $\displaystyle \frac{-1}{2}$ but am not clear on what to do after that.

However, the 3x is giving me some problems on adding and minus...

Note that 0< x <180

2. Originally Posted by Punch
Given that $\displaystyle cos 3x = 4cos^3x -3cosx$,

(a) Find the exact value of $\displaystyle 8cos^310 - 6cos10$
(b) solve the equation $\displaystyle 8cos^3x - 6cosx = - 1$ for 0< x < 180

Attempted solution

(a) 8cos^310 - 6cos10 = 2(cos3(10))
= 2cos30
= $\displaystyle \sqrt{3}$

(b) since 8cos^3x - 6cosx = 2cos 3x
2cos3x = -1
I cosine inversed $\displaystyle \frac{-1}{2}$ but am not clear on what to do after that.

However, the 3x is giving me some problems on adding and minus...

Note that 0< x <180
Do not double post.

3. Originally Posted by Prove It
Do not double post.

I am sorry this always happens because of some bugs i have in my comp..

4. Hello Punch
Originally Posted by Punch
Given that $\displaystyle cos 3x = 4cos^3x -3cosx$,

(a) Find the exact value of $\displaystyle 8cos^310 - 6cos10$
(b) solve the equation $\displaystyle 8cos^3x - 6cosx = - 1$ for 0< x < 180

Attempted solution

(a) 8cos^310 - 6cos10 = 2(cos3(10))
= 2cos30
= $\displaystyle \sqrt{3}$

(b) since 8cos^3x - 6cosx = 2cos 3x
2cos3x = -1
I cosine inversed $\displaystyle \frac{-1}{2}$ but am not clear on what to do after that.

However, the 3x is giving me some problems on adding and minus...

Note that 0< x <180
For (b), you're right so far. Note that if $\displaystyle 0^o<x<180^o$, then $\displaystyle 0^o<3x<540^o$. Then you can say, since $\displaystyle \cos3x = -\tfrac12$, we want to find all angles between $\displaystyle 0^o$ and $\displaystyle 540^o$ whose cosine is $\displaystyle -\tfrac12$.
The first of these is $\displaystyle (180-60)=120^o$, the next is $\displaystyle (180+60)=240^o$, and so on. And these are the values of $\displaystyle 3x$. So we get:
$\displaystyle 3x = 120^o, 240^o, 480^o, 600^o, ...$ (the last one being out of range.)
$\displaystyle \Rightarrow x = 40^o, 80^o, 160^o$