Question

Find the number of solution of the equation $16^{\sin^2x} + 16^{\cos^2x} = 10$, where $0 \le x \le 2\pi$.

Answer 1

8

Answer 2

Let $y = \sin^2x$. Then $\cos^2x = 1 - \sin^2x = 1 - y$. The equation becomes $16^y + 16^{1-y} = 10$. Let $z = 16^y$. Then $z + \frac{16}{z} = 10$. Multiplying by $z$, we get $z^2 + 16 = 10z$, or $z^2 - 10z + 16 = 0$. Factoring, $(z-2)(z-8) = 0$, so $z=2$ or $z=8$.

Case 1: $16^y = 2 \implies (2^4)^y = 2^1 \implies 2^{4y} = 2^1 \implies 4y=1 \implies y = \frac{1}{4}$. Since $y = \sin^2x$, we have $\sin^2x = \frac{1}{4}$, which means $\sin x = \pm \frac{1}{2}$. For $\sin x = \frac{1}{2}$, $x = \frac{\pi}{6}, \frac{5\pi}{6}$. For $\sin x = -\frac{1}{2}$, $x = \frac{7\pi}{6}, \frac{11\pi}{6}$. This gives 4 solutions.

Case 2: $16^y = 8 \implies (2^4)^y = 2^3 \implies 2^{4y} = 2^3 \implies 4y=3 \implies y = \frac{3}{4}$. Since $y = \sin^2x$, we have $\sin^2x = \frac{3}{4}$, which means $\sin x = \pm \frac{\sqrt{3}}{2}$. For $\sin x = \frac{\sqrt{3}}{2}$, $x = \frac{\pi}{3}, \frac{2\pi}{3}$. For $\sin x = -\frac{\sqrt{3}}{2}$, $x = \frac{4\pi}{3}, \frac{5\pi}{3}$. This gives another 4 solutions.

All 8 solutions are distinct and lie in the interval $0 \le x \le 2\pi$. The total number of solutions is $4 + 4 = 8$.