The equation 3 sin2 x + 10 cos x – 6 = 0 is satisfied, if (a) x = n

1.	The equation 3 sin2 x + 10 cos x – 6 = 0 is satisfied, if (a) x = nπ ± cos–1 \(\big(\frac{1}{3}\big)\) (b) x = 2nπ ± cos–1 \(\big(\frac{1}{3}\big)\) (c) x = nπ ± cos–1 \(\big(\frac{1}{6}\big)\) (d) x = 2nπ ± cos–1 \(\big(\frac{1}{6}\big)\)
Answer» Answer: (b) 2nπ ± cos^–1\(\big(\frac{1}{3}\big)\) 3 sin² x + 10 cos x – 6 = 0 ⇒ 3 (1 – cos² x) + 10 cos x – 6 = 0 ⇒ – 3 cos²x + 10 cos x – 3 = 0 ⇒ (cos x – 3) (1 – 3 cos x) = 0 ⇒ cos x = 3 or cos x = \(\frac{1}{3}\) Since cos x = 3 is inadmissible, therefore, cos x = \(\frac{1}{3}\) ⇒ x = cos^–1\(\big(\frac{1}{3}\big)\) The general solution is x = 2nπ ± cos^–1\(\big(\frac{1}{3}\big)\)

The equation 3 sin2 x + 10 cos x – 6 = 0 is satisfied, if (a) x = nπ ± cos–1 \(\big(\frac{1}{3}\big)\) (b) x = 2nπ ± cos–1 \(\big(\frac{1}{3}\big)\) (c) x = nπ ± cos–1 \(\big(\frac{1}{6}\big)\) (d) x = 2nπ ± cos–1 \(\big(\frac{1}{6}\big)\)

Answer»

Answer: (b) 2nπ ± cos^–1\(\big(\frac{1}{3}\big)\)

3 sin² x + 10 cos x – 6 = 0

⇒ 3 (1 – cos² x) + 10 cos x – 6 = 0

⇒ – 3 cos²x + 10 cos x – 3 = 0

⇒ (cos x – 3) (1 – 3 cos x) = 0

⇒ cos x = 3 or cos x = \(\frac{1}{3}\)

Since cos x = 3 is inadmissible, therefore, cos x = \(\frac{1}{3}\)

⇒ x = cos^–1\(\big(\frac{1}{3}\big)\)

The general solution is

x = 2nπ ± cos^–1\(\big(\frac{1}{3}\big)\)