Verified Solution • 2025–26 Session moderate

QJEEM21Ma7877: JEE Main 2021 Mathematics - Class 12 JEEM21Ma7877 Solution: The solutions of the equation

The solutions of the equation |1+sin2xsin2xsin2xcos2x1+cos2xcos2x4sin⁡2x4sin⁡2x1+4sin⁡2x|=0,(0<x<π), are

Multiple Choice Options

  • A π12,π6{\pi \over {12}},{\pi \over 6}
  • B π6,5π6{\pi \over 6},{{5\pi } \over 6}
  • C 5π12,7π12{{5\pi } \over {12}},{{7\pi } \over {12}}
  • D 7π12,11π12{{7\pi } \over {12}},{{11\pi } \over {12}}

Step-by-Step Verified Solution

By using C1 →\to C1 −- C2 and C3 →\to C3 −- C2 we get

|1sin2x0−11+cos2x−104sin⁡2x1|=0

Expanding by R1 we get

1(1+cos⁡2x+4sin⁡2x)−sin⁡2x(−1)=01(1 + {\cos ^2}x + 4\sin 2x) - {\sin ^2}x( - 1) = 0

⇒2+4sin⁡2x=0\Rightarrow 2 + 4\sin 2x = 0

⇒sin⁡2x=−12\Rightarrow \sin 2x = {{ - 1} \over 2}

⇒2x=nπ+(−1)n(−π6),n∈Z\Rightarrow 2x = n\pi + {( - 1)^n}\left( {{{ - \pi } \over 6}} \right),n \in Z

∴\therefore 2x=7π6,11π62x = {{7\pi } \over 6},{{11\pi } \over 6}

⇒x=7π12,11π2\Rightarrow x = {{7\pi } \over {12}},{{11\pi } \over 2}

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