2sin²x+4sin²x·cos²x=2. 2sin²x(1+2cos²x)=2. sin²x(1+2cos²x)=1. Let sin²x=t: t(1+2(1-t))=1, t(3-2t)=1, 2t²-3t+1=0, (2t-1)(t-1)=0. t=1/2 or t=1. sin²x=1/2 gives 4 solutions, sin²x=1 gives 2 solutions. Total=6
The number of solutions of 2sin²x + sin²2x = 2 in [0, 2π] is:
JEE Main 2026 — Mathematics Trigonometry
The number of solutions of 2sin²x + sin²2x = 2 in [0, 2π] is:
Verified 30 May 2026.
Options
- A
6
- B
4
- C
8
- D
2
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