If 1+2(\sin x+\cos x)(\sin x-\cos x)=0 where 0 \lt x \lt 360^{\circ} then how many values does x take?
- A. Only one value
- B. Only two values
- C. Only three values
- D. Four values ✓
Correct Answer: D. Four values
Explanation
Simplify the equation: 1 + 2(\sin^2 x - \cos^2 x) = 0 \implies 1 - 2\cos 2x = 0 \implies \cos 2x = \frac{1}{2}. For x \in (0^{\circ}, 360^{\circ}), 2x \in (0^{\circ}, 720^{\circ}). The equation \cos 2x = \frac{1}{2} has 4 solutions in this interval (at 2x = 60^{\circ}, 300^{\circ}, 420^{\circ}, 660^{\circ}). Therefore, x takes 4 values.