# Angles of a triangle

1. May 14, 2013

### Saitama

1. The problem statement, all variables and given/known data
In a $\Delta ABC$, angle A is greater than angle B. If the measures of A and B satisfy the equation $3\sin x-4\sin^3 x-0.75=0$, then angle C is equal to
A)$\pi/3$
B)$\pi/2$
C)$2\pi/3$
D)$5\pi/6$

2. Relevant equations

3. The attempt at a solution
The given equation can be rewritten as $12\sin x-16\sin^3 x-3=0$. This is a cubic equation in $\sin x$, how am I supposed to solve this?

Any help is appreciated. Thanks!

2. May 14, 2013

### SteamKing

Staff Emeritus
Let y = sin x, then you have a cubic in y.

You could also try manipulating one of the trig identities for sine of the multiple of an angle.

3. May 14, 2013

### Staff: Mentor

4. May 14, 2013

### Saitama

Sorry but I am still lost, how am I supposed to solve $16y^3-12y+3=0$?

5. May 14, 2013

### davidchen9568

6. May 14, 2013

### Saitama

Cool!

I totally forgot about that.

That gives $\sin 3x=3/4$, hence
$$x=\frac{1}{3}\arcsin\frac{3}{4}, \frac{1}{3}\left(\pi-\arcsin\frac{3}{4}\right)$$

$C=\pi-(A+B)$ where $A+B=\pi/3$, therefore $C=2\pi/3$.

Thank you David!

No, I would never use the alternative method, that isn't in my syllabus and you already gave such a nice method. :tongue2:

7. May 14, 2013

### Curious3141

It's interesting that the 0.75 doesn't matter a whit here in solving it.