Solve cos x – √3 sin x = 1, 0° < x < 360°

1.	Solve cos x – √3 sin x = 1, 0° < x < 360°
Answer» Dividing both the sides of the equation cos x − √3 sin x = 1 by \(\sqrt{(1)^2 +(- \sqrt{3})^2}\) = 2 , we get \(\frac{1}{2}\) cos x - \(\frac{\sqrt{3}}{2}\) sin x = \(\frac{1}{2}\) ⇒ cos 60° cos x – sin 60° sin x = \(\frac{1}{2}\) ⇒ cos (x + 60°) = cos 60° ⇒ cos (x + 60°) = cos 60° = cos (360° – 60°) = cos (360° + 60°) ⇒ x + 60° = 60° or 300° or 420° ⇒ x = 0°, 240°, 360°.

Answer»

Dividing both the sides of the equation cos x − √3 sin x = 1 by

\(\sqrt{(1)^2 +(- \sqrt{3})^2}\) = 2 , we get

\(\frac{1}{2}\) cos x - \(\frac{\sqrt{3}}{2}\) sin x = \(\frac{1}{2}\)

⇒ cos 60° cos x – sin 60° sin x = \(\frac{1}{2}\)

⇒ cos (x + 60°) = cos 60°

⇒ cos (x + 60°) = cos 60° = cos (360° – 60°) = cos (360° + 60°)

⇒ x + 60° = 60° or 300° or 420°

⇒ x = 0°, 240°, 360°.

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