How many solutions does $6 \sin ^{2} x-\cos ^{2} x=3$ have for $0 \leq x \leq 2 \pi$ ?
A) 4
B) 0
C) 1
D) none of the above.

Question

Quizplus · Accepted Answer

The equation can be rewritten using the identity $\cos^2 x = 1 - \sin^2 x$ as $6 \sin^2 x - (1 - \sin^2 x) = 3$. Simplifying gives $7 \sin^2 x = 4$, so $\sin^2 x = \frac{4}{7}$. Thus, $\sin x = \pm \sqrt{\frac{4}{7}}$. In the interval $0 \leq x \leq 2\pi$, each value of $\sin x$ corresponds to two solutions (one in each half of the unit circle), resulting in 4 solutions in total.

How Many Solutions Does 6sin⁡2x−cos⁡2x=36 \sin ^{2} x-\cos ^{2} x=36sin2x−cos2x=3 Have For

How Many Solutions Does $6 \sin ^{2} x-\cos ^{2} x=3$ Have For