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QJEEM21Mae8af: JEE Main 2021 Mathematics - Class 11 JEEM21Mae8af Solution: Let S be the sum of all solutions (in radians) of the equation…

Let S be the sum of all solutions (in radians) of the equation sin⁡4θ+cos⁡4θ−sin⁡θcos⁡θ=0{\sin ^4}\theta + {\cos ^4}\theta - \sin \theta \cos \theta = 0 in [0, 4 π\pi ]. Then 8Sπ{{8S} \over \pi } is equal to ____________.

Step-by-Step Verified Solution

Given equation

sin⁡4θ+cos⁡4θ−sin⁡θcos⁡θ=0{\sin ^4}\theta + {\cos ^4}\theta - \sin \theta \cos \theta = 0

⇒1−sin⁡2θcos⁡2θ−sin⁡θcos⁡θ=0\Rightarrow 1 - {\sin ^2}\theta {\cos ^2}\theta - \sin \theta \cos \theta = 0

⇒2−(sin⁡2θ)2−sin⁡2θ=0\Rightarrow 2 - {(\sin 2\theta )^2} - \sin 2\theta = 0

⇒(sin⁡2θ)2+(sin⁡2θ)−2=0\Rightarrow {(\sin 2\theta )^2} + (\sin 2\theta ) - 2 = 0

⇒(sin⁡2θ+2)(sin⁡2θ−1)=0\Rightarrow (\sin 2\theta + 2)(\sin 2\theta - 1) = 0

⇒sin⁡2θ=1\Rightarrow \sin 2\theta = 1 or sin⁡2θ=−2\sin 2\theta = - 2 (Not Possible)

⇒2θ=π2,5π2,9π2,13π2\Rightarrow 2\theta = {\pi \over 2},{{5\pi } \over 2},{{9\pi } \over 2},{{13\pi } \over 2}

⇒θ=π4,5π4,9π4,13π4\Rightarrow \theta = {\pi \over 4},{{5\pi } \over 4},{{9\pi } \over 4},{{13\pi } \over 4}

⇒S=π4+5π4+9π4+13π4=7π\Rightarrow S = {\pi \over 4} + {{5\pi } \over 4} + {{9\pi } \over 4} + {{13\pi } \over 4} = 7\pi

⇒8Sπ=8×7ππ=56.00\Rightarrow {{8S} \over \pi } = {{8 \times 7\pi } \over \pi } = 56.00

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