Discussion Forum

 The number of distinct solutions of the equation $\frac{5}{4}\cos^{2}2x+\cos^{4}x+\sin^{4}x+\cos^{6}x+\sin^{6}x=2$ in the interval [0,2π) is 

Answer:

Explanation:

Here,     $\frac{5}{4}\cos^{2}2x+(\cos^{4}x+\sin^{4}x)+(\cos^{6}x+\sin^{6}x)=2$

               $\Rightarrow \frac{5}{4}\cot^{}2x+[(\cos^{2}x+\sin^{2}x)^{2}-2\sin^{2}x\cos^{2}x]$

                            $+(\cos^{2}x+\sin^{2}x)[(\cos^{2}x+\sin^{2}x)^{2}-3\sin^{2}x\cos^{2}x]=2$

             $\Rightarrow \frac{5}{4}\cos^{2}2x+(1-2\sin^{2}x\cos^{2}x)+(1-3\cos^{2} x \sin^{2}x)=2$

          $\Rightarrow \frac{5}{4}\cos^{2}2x-5\sin^{2}x\cos^{2}x=0$

  $\Rightarrow$                $\frac{5}{4}\cos^{2}2x-\frac{5}{4}\sin^{2}2x=0$

 $\Rightarrow$                   $\frac{5}{4}\cos^{2}2x-\frac{5}{4}+\frac{5}{4}\cos^{2}2x=0$

 $\Rightarrow$           $\frac{5}{2}\cos^{2}2x=\frac{5}{4}$

 $\Rightarrow$            $\cos^{2}2x=\frac{1}{2}\Rightarrow 2\cos^{2}2x=1$

 $\Rightarrow$             1+cos 4x=1

 $\Rightarrow$     cos 4x=0, as  $0\leq x\leq 2\pi$

  $\therefore$

             $4x=\left\{\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2},\frac{9\pi}{2},\frac{11\pi}{2},\frac{13\pi}{2},\frac{15\pi}{2}\right\}$

            as   $0\leq 4x\leq8\pi$

    $\Rightarrow x=\left\{\frac{\pi}{8},\frac{3\pi}{8},\frac{5\pi}{8},\frac{7\pi}{8},\frac{9\pi}{8},\frac{11\pi}{8},\frac{13\pi}{8},\frac{15\pi}{8}\right\}$

             Hence, the total number of solutions is 8.