If n is the number of solutions of the equation 2 cosx(4 sin( π 4+x) sin( π 4-x)-1)=1, x∈ [0,π ] and S is the sum of all these solutions, then the…

$$2\mathrm{cos}x(4\mathrm{sin}(\frac{\pi }{4}+x)\mathrm{sin}(\frac{\pi }{4}-x)-1)=1$$2cosx(4sin(4π​+x)sin(4π​−x)−1)=1

$$\Rightarrow 2\mathrm{cos}x(2\mathrm{cos}(2x)-2\mathrm{cos}(\frac{\pi }{2})-1)=1$$⇒2cosx(2cos(2x)−2cos(2π​)−1)=1

$$\Rightarrow 2\mathrm{cos}x(2.(2{\mathrm{cos}}^{2}x-1)-1)=1$$⇒2cosx(2.(2cos2x−1)−1)=1

$$\Rightarrow 2\mathrm{cos}x(4{\mathrm{cos}}^{2}x-3)=1$$⇒2cosx(4cos2x−3)=1

$$\Rightarrow 4{\mathrm{cos}}^{3}x-3\mathrm{cos}x=\frac{1}{2}$$⇒4cos3x−3cosx=21​

$$\Rightarrow \mathrm{cos}3x=\frac{1}{2}$$⇒cos3x=21​

$$\Rightarrow 3x=\frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3}$$⇒3x=3π​,35π​,37π​

$$\Rightarrow x=\frac{\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9}$$⇒x=9π​,95π​,97π​

No. of solutions $=n=3$=n=3

Sum of solutions $=S=\frac{\pi }{9}+\frac{5\pi }{9}+\frac{7\pi }{9}=\frac{13\pi }{9}$=S=9π​+95π​+97π​=913π​