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COMEDK · Maths · 9. Trigonometric Equations

The general solution of \(2 \cos 4 x+\sin ^2 2 x=0\) is

  1. A \(x=\frac{n \pi}{2} \pm \sin ^{-1}\left(\frac{1}{5}\right)\)
  2. B \(x=\frac{n \pi}{4}+\frac{(-1)^n}{4} \sin ^{-1}\left( \pm \frac{2 \sqrt{2}}{3}\right)\)
  3. C \(x=\frac{n \pi}{2} \pm \cos ^{-1}\left(\frac{1}{5}\right)\)
  4. D \(x=\frac{n \pi}{4}+\frac{(-1)^n}{4} \cos ^{-1}\left(\frac{1}{5}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(x=\frac{n \pi}{4}+\frac{(-1)^n}{4} \sin ^{-1}\left( \pm \frac{2 \sqrt{2}}{3}\right)\)

Step-by-step Solution

Detailed explanation

Given, \(2 \cos 4 x+\sin ^2 2 x=0\) \(\begin{array}{ll} \Rightarrow & 2 \cos 4 x+\left(\frac{1-\cos 4 x}{2}\right)=0 \\ \Rightarrow & 3 \cos 4 x+1=0 \end{array}\)…