Trigonometrical Equations
hard

यदि समीकरण $2 \cos x\left(4 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)-1\right)=1$, $x \in[0, \pi]$ के हलों की संख्या $n$ है तथा $S$ इन सभी हलों का योगफल है, तब क्रमित युग्म $( n , S )$ है

A

$(3,13 \pi / 3)$

B

$(2,2 \pi / 3)$

C

$(2,8 \pi / 9)$

D

$(3,5 \pi / 3)$

(JEE MAIN-2021)

Solution

$2 \cos x\left(4 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)-1\right)=1$

$2 \cos x\left(4\left(\sin ^{2} \frac{\pi}{4}-\sin ^{2} x\right)-1\right)=1$

$2 \cos x\left(4\left(\frac{1}{2}-\sin ^{2} x\right)-1\right)=1$

$2 \cos x\left(2-4 \sin ^{2} x-1\right)=1$

$2 \cos x\left(1-4 \sin ^{2} x\right)=1$

$2 \cos x\left(4 \cos ^{2} x-3\right)=1$

$4 \cos ^{3} x-3 \cos x=\frac{1}{2}$

$\cos 3 x=\frac{1}{2}$

$\mathrm{x} \in[0, \pi] \therefore 3 \mathrm{x} \in[0,3 \pi]$

Standard 11
Mathematics

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