Find the smallest positive value of x satisfying the equation logcos x

1.	Find the smallest positive value of x satisfying the equation logcos x sin x + logsin x cos x = 2
Answer» Let log_{cos x} sin x = p. Then, log_{sin x} cos x = \(\frac{1}{p}\). ∴ log _{cos x} sin x + log_{sin x} cos x = 2 ⇒ p + \(\frac{1}{p}\) = 2 ⇒ p² – 2p + 1 = 0 ⇒ (p – 1)² = 0 ⇒ p = 1 ⇒ log_{cos x} sin x = 1 ⇒ sin x = cos x The smallest positive value of x for which sin x = cos x is x = \(\frac{\pi}{4}\) .

Find the smallest positive value of x satisfying the equation logcos x sin x + logsin x cos x = 2

Answer»

Let log_{cos x} sin x = p. Then, log_{sin x} cos x = \(\frac{1}{p}\).

∴ log _{cos x} sin x + log_{sin x} cos x = 2

⇒ p + \(\frac{1}{p}\) = 2

⇒ p² – 2p + 1 = 0

⇒ (p – 1)² = 0

⇒ p = 1

⇒ log_{cos x} sin x = 1

⇒ sin x = cos x

The smallest positive value of x for which sin x = cos x is x = \(\frac{\pi}{4}\) .