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  • If is the number of solutions of the equation <img alt="2 \cos x\left(4 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)-1\right)=1$, $x \in[0, \pi]" src="https://ent

If n is the number of solutions of the equation 2 \cos x\left(4 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)-1\right)=1$, $x \in[0, \pi] and S is the sum of all these solutions, then the ordered pair (\mathrm{n}, \mathrm{S}) is
Option: 1 (2,8 \pi / 9)
Option: 2 (3,13 \pi / 9)
Option: 3 (2,2 \pi / 3)
Option: 4 (3,5 \pi / 3)

Answers (1)

best_answer

2\cos x\left [ 4\left ( \sin ^{2} \frac{\pi }{4}-\sin ^{2}x\right ) -1\right ]= 1
\Rightarrow 2\cos x\left [ 4\left ( \frac{1}{2} -\sin ^{2}x\right )-1\right ]= 1
\Rightarrow 2\cos x\left [2-4\sin ^{2}x-1\right ]= 1
\Rightarrow 2\cos x\left [1-4\sin ^{2}x\right ]= 1
\Rightarrow 2\cos x\left [1-4+4\cos ^{2}x\right ]= 1
\Rightarrow 2\cos x\left [4\cos ^{2}x-3\right ]= 1
\Rightarrow 2\left ( 4\cos ^{3}x-3\cos x \right )= 1
\Rightarrow \cos 3x= \frac{1}{2}
\Rightarrow 3x= \frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3}
\Rightarrow x= \frac{\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9}
\Rightarrow n= 3,S= \frac{13\pi }{9}

Posted by

Kuldeep Maurya

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