# Doubt in trigonometry

Yes

See my first vision(hvnt done in copy) telling looking like binomial application because terms are cancelling

Okay I got it,thanks

Use the fact that if a+b+c=0 then

\left(\dfrac{a^5+b^5+c^5}{5} \right) = \left(\dfrac{a^3+b^3+c^3}{3} \right) \left(\dfrac{a^2+b^2+c^2}{2} \right)

I did the same

I used this

If a+b+c=0 then a^5+b^5+c^5=-3abc(ab+bc+ac)

Here for this question (ab+bc+ac)=-\frac{3}{4}

So we need to only focus on -3abc term

In fact with \alpha= x, \beta = x+\dfrac{2\pi}{3}, \gamma = x+\dfrac{4\pi}{3}, we have that \alpha, \beta, \gamma are roots of \cos 3\theta = \cos 3x or 4 \cos^3\theta -3 \cos \theta -\cos 3x=0

so that p=\cos \alpha, q=\cos \beta, r=\cos \gamma are roots of 4y^3-3y-\cos 3x=0. Hence p+q+r=0, \sum pq=-\dfrac{3}{4}, pqr=\dfrac{\cos 3x}{4}

So, 4p^3-3p-\cos 3x=0 \Rightarrow 4p^5-3p^3-\cos 3x (p^2)=0 \Rightarrow 4\sum p^5-3\sum p^3-\cos 3x \sum (p^2)=0

Thus 4\sum p^5=3\sum p^3+\cos 3x \sum (p^2)

Since p+q+r=0, \sum p^3= 3pqr=\dfrac{3\cos 3x}{4},\cos 3x \sum p^2=-2\cos 3x \sum pq =\dfrac{-3\cos 3x}{2}

Thus the equation becomes \cos 3x=0