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

Concept:
Use \[ \cos^2 x = 1-\sin^2 x \] and reduce the equation to a quadratic form. ip

Step 1: Set \(t=\sin^2 x\).
Then \[ \cos^2 x = 1-t \] So the equation becomes: \[ 16^t + 16^{1-t}=10 \] ip

Step 2: Convert into a quadratic equation.
Let \[ u=16^t \] Then \[ 16^{1-t}=\frac{16}{u} \] So, \[ u+\frac{16}{u}=10 \] \[ u^2-10u+16=0 \] \[ (u-8)(u-2)=0 \] Thus, \[ u=8 \quad \text{or} \quad u=2 \] ip

Step 3: Find the corresponding values of \(\sin^2 x\).
If \[ 16^t=8=2^3 \] then \[ 2^{4t}=2^3 \Rightarrow 4t=3 \Rightarrow t=\frac34 \] If \[ 16^t=2=2^1 \] then \[ 2^{4t}=2^1 \Rightarrow 4t=1 \Rightarrow t=\frac14 \] So, \[ \sin^2 x=\frac14 \quad \text{or} \quad \sin^2 x=\frac34 \] ip

Step 4: Count solutions in \([0,2\pi]\).
\[ \sin^2 x=\frac14 \Rightarrow \sin x=\pm \frac12 \] gives \(4\) solutions in \([0,2\pi]\). Also, \[ \sin^2 x=\frac34 \Rightarrow \sin x=\pm \frac{\sqrt3}{2} \] also gives \(4\) solutions. Total number of solutions: \[ 4+4=8 \] ip Hence, the correct answer is:
\[ \boxed{(A)\ 8} \]