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JEE Mains · Maths · STD 11 - Trigonometrical equations

The number of  \(x \in  [0,2\pi ]\)  for which \(\left| {\sqrt {2\,{{\sin }^4}\,x\, + \,18\,{{\cos }^2}\,x}  - \,\sqrt {2\,{{\cos }^4}\,x\, + \,18\,{{\sin }^2}\,x} } \right| = 1\) is

  1. A \(2\)
  2. B \(6\)
  3. C \(4\)
  4. D \(8\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(8\)

Step-by-step Solution

Detailed explanation

\(|\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}-\sqrt{2 \cos ^{4} x+18 \sin ^{2} x}|=1\) \(\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}-\sqrt{2} \cos ^{4} x+18 \sin ^{2} x=\pm 1\) \(\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}=\pm 1+\sqrt{2 \cos ^{4} x+18 \sin ^{2} x}\) by squaring both the sides we will…