# Solve the equation: $\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1$

How to solve the equation

$$\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1$$

Can anyone give me some hints?

• Use the addition angle formulae for the sine and cosine functions. Commented Dec 24, 2015 at 18:42
• Hint: If you know the complex form $2cos(a) = z+\frac{1}{z}$ where $z=e^{ia}$ then you get the polynomial equation $1+z+z^2+z^3+z^4+z^5+z^6=-z^3$.
– Paul
Commented Dec 24, 2015 at 18:56
• @Dr.MV Can you suggest a solution for this question? Because I saw your answers from other posts are quite good Commented Dec 24, 2015 at 19:14
• @Mathxx Thank you for your compliment! That made my day! Happy Holidays! - Mark Commented Dec 24, 2015 at 19:55

HINT:

$$\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1\Longleftrightarrow$$ $$1+6\cos^2(x)-20\cos^4(x)+16\cos^6(x)=1\Longleftrightarrow$$

Substitute $y=\cos^2(x)$:

$$16y^3-20y^2+6y=0\Longleftrightarrow$$ $$2y(2y-1)(4y-3)=0\Longleftrightarrow$$ $$y(2y-1)(4y-3)=0$$

Hint: Simplify $1+6\cos^2(x)-20\cos^4(x)+16\cos^6(x)=1$ and then substitute $y=\cos^2(x)$ and then you will get $16y^3-20y^2+6y+1=1$

HINT: $$\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1$$ $$\cos^2(x)+\cos^2(3x)=1-\cos^2(2x)$$ $$\cos^2(x)+\cos^2(3x)=(1-\cos(2x))(1+\cos(2x))$$ $$\cos^2(x)+\cos^2(3x)=(2\sin^2(x))(2\cos^2(x))$$ $$\cos^2(x)+(4\cos^3(x)-3\cos (x))^2=2(1-\cos^2(x))(2\cos^2(x))$$ $$16\cos^6x-20\cos^4x+6\cos^2x=0$$ now, let $\cos^2 x=t$, $$8t^2-10t+3t=0$$ $$t(2t-1)(4t-3)=0$$

$$\cos^2(x)+\cos^2(2x)+\cos^2(3x)-1$$ $$=\cos^2(x)-\sin^23x+\cos^2(2x)$$ $$=\cos(3x-x)\cos(3x+x)+\cos^2(2x)$$ $$=\cos2x(\cos4x+\cos2x)$$

Now use Prosthaphaeresis Formulas on $$\cos4x+\cos2x$$

Should I use a single word more?

Alternatively using $\cos2A=2\cos^2A-1,$

$$\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1$$

$$\iff\cos2x+\cos4x+\cos6x+1=0$$

$$\cos6x=2\cos^23x-1$$ and use Prosthaphaeresis Formula on $$\cos2x+\cos4x$$

$\cos^2(x)+[2 \cos^2x-1]^2 +[4 \cos^3 x-3 \cos x]^2 = 1$, Simplify the equation to get: $\cos^2 x [8 \cos^4 x - 10 \cos^2 x + 3] = 0$. Hope this will help.