Question

find general solution for cos4x=cos2x​

Answer 1

$\underline{\large{\sf Answer:}}$

we have given equation,

$\sf cos4x = cos2x$

$\sf cos4x - cos2x = 0$

using the formula,

$\sf cos C - cosD = -2sin(\frac{C+D}{2})sin(\frac{C-D}{2})$

Therefore,

$\sf cos4x - cos2x = -2sin(\frac{4x+2x}{2}).sin(\frac{4x-2x}{2})=0$

$\implies \sf -2sin(\frac{6x}{2}).sin(\frac{2x}{2})=0$

$\implies\sf-2sin3x.sinx=0$

∴ sin3x = 0

OR

∴ sinx = 0

we know,

sinΦ = 0 implies Φ = nπ , n€Z

∴ 3x = nπ OR x = mπ

Hence, the general solution is $\sf x = \frac{n\pi}{3}$ OR $\sf x =m \pi$ ,here n, m € Z.