Reduced Value of Cosine

As $(1-\cos x)(1+\cos x)=\sin^2x=\dfrac{1-\cos2x}2, 1+\cos x=\dfrac{1-\cos2x}{2(1-\cos x)}$ if $1-\cos x\ne0$

Set $x=\theta, 2\theta, 4\theta$ one by one to find $1-\cos8\theta=1-\cos\theta\implies \cos8\theta=\cos\theta\implies8\theta=360^\circ n \pm\theta$ where n is any integer

Taking the '+' sign, $7\theta=360^\circ n\iff \theta=\dfrac{360^\circ n}7$

Now we need $0<\dfrac{360^\circ n}7<90\iff 0< n<\dfrac74\implies n=1$ as n is an integer

Similarly deal with '-' sign to find 2 more solutions