Solve tan 3x = cot 5x (0 < x < 2π).

1.	Solve tan 3x = cot 5x (0 < x < 2π).
Answer» tan 3x = tan \(\big(\) \(\frac{\pi}{2}\) - 5x \(\big)\) ⇒ 3x = nπ + \(\big(\) \(\frac{\pi}{2}\) - 5x \(\big)\) (∵ tan θ = tan α ⇒ θ = nπ + α) ⇒ 8x = (2n + 1)\(\frac{\pi}{2}\) ⇒ x = (2n + 1) \(\frac{\pi}{16}\) ∴ Putting n = 0, 1, 2, ..... 15, we see that the values of x between 0 and 2π are x = \(\frac{\pi}{16}\) ,\(\frac{3\pi}{16}\) , \(\frac{5\pi}{16}\) ,..., \(\frac{31\pi}{16}\).

Answer»

tan 3x = tan \(\big(\) \(\frac{\pi}{2}\) - 5x \(\big)\)

⇒ 3x = nπ + \(\big(\) \(\frac{\pi}{2}\) - 5x \(\big)\) (∵ tan θ = tan α ⇒ θ = nπ + α)

⇒ 8x = (2n + 1)\(\frac{\pi}{2}\) ⇒ x = (2n + 1) \(\frac{\pi}{16}\)

∴ Putting n = 0, 1, 2, ..... 15, we see that the values of x between 0 and 2π are

x = \(\frac{\pi}{16}\) ,\(\frac{3\pi}{16}\) , \(\frac{5\pi}{16}\) ,..., \(\frac{31\pi}{16}\).

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