Question Number 194315 by cortano12 last updated on 03/Jul/23
$$\:\:\:\: \\ $$$$ \mathrm{4sin}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{x}\:+\mathrm{1}\:=\:\mathrm{2cos}\:\mathrm{4x}\: \\ $$$$\: \\ $$
Answered by Frix last updated on 03/Jul/23
![This can be transformed to [s=sin x]: s^4 +(s^3 /2)−s^2 −(s/2)+(1/(16))=0 which can be solved the usual way.](Q194324.png)
$$\mathrm{This}\:\mathrm{can}\:\mathrm{be}\:\mathrm{transformed}\:\mathrm{to}\:\left[{s}=\mathrm{sin}\:{x}\right]: \\ $$$${s}^{\mathrm{4}} +\frac{{s}^{\mathrm{3}} }{\mathrm{2}}−{s}^{\mathrm{2}} −\frac{{s}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{16}}=\mathrm{0} \\ $$$$\mathrm{which}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{the}\:\mathrm{usual}\:\mathrm{way}. \\ $$