The number of distinct solutions of the equation, log21∣sinx∣=2−log21∣cosx∣ in the interval [0,2π], is ______.
Answer: 8
Solution
Combining the two logarithms gives log1/2∣sinxcosx∣=2, so ∣sinxcosx∣=(21)2=41. Using sin2x=2sinxcosx this becomes ∣sin2x∣=21. As x ranges over [0,2π], 2x ranges over [0,4π], and there ∣sin2x∣=21 has four solutions from sin2x=21 and four from sin2x=−21, giving eight values of x in all.Graph showing the 8 solutions in [0, 2π]
JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Integer answer