Solving $\cos^2 x+\cos^2 2x +\cos^2 3x=1$

Use double angles to get the equivalent equation $$\cos(0x) + \cos(2x) + \cos(4x) + \cos(6x) = 0.$$

Pair (first and last term) and (middle ones) to get the equivalent form $$\cos(x)\cos(2x)\cos(3x) = 0.$$


Use double angle identify to write the equation as

$$\cos 2x+\cos 4x +\cos 6x+1=0$$

Let $t=\cos 2x$. Then, the equation is

$$2t^3+t^2-t=0\implies t(t+1)(2t-1)=0$$

Thus, $\cos 2x= 0,\> -1, \> \frac12$ and the solutions are $x = \frac\pi4+\frac{n\pi}2,\> \frac\pi2+ n\pi,\> \pm\frac\pi6+\frac{2n\pi}3$.


Let $t:=\cos^2x$.

We have

$$t+(2t-1)^2+(4t-3)^2t=1$$

or

$$16t^3-20t^2+6t=0.$$

The roots are $0,\dfrac12,\dfrac34$, nothing really difficult.

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Trigonometry

Algebra Precalculus

Contest Math

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