COMEDK · Maths · 9. Trigonometric Equations
The general solution of \(2 \cos 4 x+\sin ^2 2 x=0\) is
- A \(x=\dfrac{n \pi}{2} \pm \sin ^{-1}\left(\dfrac{1}{5}\right)\)
- B \(x=\dfrac{n \pi}{4}+\dfrac{(-1)^n}{4} \cos ^{-1}\left(\dfrac{1}{5}\right)\)
- C \(x = \frac{n\pi}{2} \pm \frac{1}{4}\cos^{-1}\left(-\frac{1}{3}\right)\)
- D \(x=\dfrac{n \pi}{2} \pm \cos ^{-1}\left(\dfrac{1}{5}\right)\)
Answer & Solution
Correct Answer
(C) \(x = \frac{n\pi}{2} \pm \frac{1}{4}\cos^{-1}\left(-\frac{1}{3}\right)\)
Step-by-step Solution
Detailed explanation
\(2\cos 4x + \sin^2 2x = 0\) Using \(\cos 4x = 1 - 2\sin^2 2x\) \(\Rightarrow\) \(\sin^2 2x = \dfrac{1 - \cos 4x}{2}\): \(2\cos 4x + \dfrac{1 - \cos 4x}{2} = 0\) \(4\cos 4x + 1 - \cos 4x = 0\) \(3\cos 4x = -1\) \(\cos 4x = -\dfrac{1}{3}\) General solution:…
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