Question Number 213504 by Spillover last updated on 06/Nov/24
Answered by A5T last updated on 06/Nov/24
$${psin}^{\mathrm{2}} {x}−{q}\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)={p}−{q} \\ $$$${psin}^{\mathrm{2}} {x}−{q}+{qsin}^{\mathrm{2}} {x}={p}−{q} \\ $$$${sin}^{\mathrm{2}} {x}\left({p}+{q}\right)={p}\Rightarrow{sin}^{\mathrm{2}} {x}=\frac{{p}}{{p}+{q}}\Rightarrow{sin}^{\mathrm{4}} {x}=\frac{{p}^{\mathrm{2}} }{\left({p}+{q}\right)^{\mathrm{2}} } \\ $$$${cos}^{\mathrm{4}} {x}=\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} =\left(\mathrm{1}−\frac{{p}}{{p}+{q}}\right)^{\mathrm{2}} =\frac{{q}^{\mathrm{2}} }{\left({p}+{q}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{sin}^{\mathrm{4}} {x}}{{p}}+\frac{{cos}^{\mathrm{4}} {x}}{{q}}=\frac{{p}}{\left({p}+{q}\right)^{\mathrm{2}} }+\frac{{q}}{\left({p}+{q}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{p}+{q}} \\ $$
Commented by Spillover last updated on 07/Nov/24
$${great}\:{solution}.{thanks} \\ $$