Question

The number of solutions of \(16^{\sin^2 x} + 16^{\cos^2 x} = 10\) in \(0 \le x \le 2\pi\) are

• A. \(8\)
• B. \(10\)
• C. \(6\)
• D. \(4\)

Hint:

Whenever both \(\sin^2 x\) and \(\cos^2 x\) appear, substituting \(t=\sin^2 x\) is often the cleanest approach.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We convert the exponential trigonometric equation into a quadratic form using the identity \(\cos^2 x = 1 - \sin^2 x\).
Step 2: Key Formula or Approach:
Let \(16^{\sin^2 x} = t\). Then \(16^{\cos^2 x} = 16^{1-\sin^2 x} = 16/t\).
Step 3: Detailed Explanation:
The equation becomes:
\[ t + \frac{16}{t} = 10 \implies t^2 - 10t + 16 = 0 \] Factorize: \((t-8)(t-2) = 0 \implies t = 8\) or \(t = 2\).
Case I: \(16^{\sin^2 x} = 8 \implies (2^4)^{\sin^2 x} = 2^3 \implies 4 \sin^2 x = 3 \implies \sin^2 x = 3/4\).
\(\sin x = \pm \sqrt{3}/2\). In \([0, 2\pi]\), this gives 4 solutions: \(\pi/3, 2\pi/3, 4\pi/3, 5\pi/3\).
Case II: \(16^{\sin^2 x} = 2 \implies (2^4)^{\sin^2 x} = 2^1 \implies 4 \sin^2 x = 1 \implies \sin^2 x = 1/4\).
\(\sin x = \pm 1/2\). In \([0, 2\pi]\), this gives 4 solutions: \(\pi/6, 5\pi/6, 7\pi/6, 11\pi/6\).
Total solutions \(= 4 + 4 = 8\).
Step 4: Final Answer:
The number of solutions is \(8\).