Solving a Triangle's Angles with an Equation

[Saitama]

Homework Statement

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)$5\pi/6$

Homework Equations

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!

 

[SteamKing]

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.

 

[NascentOxygen]

After solving this you could check your result with wolfram alpha: http://m.wolframalpha.com/input/?i=Solve++12sin%28x%29-16%28sin%28x%29%29^3+-3%3D0&x=2&y=5&js=off

 

[Saitama]

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.

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

 

[davidchen9568]

You may need this formula:
$$3sinx-4{ sin }^{ 3 }x=sin3x$$

Edit: It is possible to solve it directly using Cardano's method:
https://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method

 

[Saitama]

You may need this formula:
$$3sinx-4{ sin }^{ 3 }x=sin3x$$

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:

 

[Curious3141]

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