Question

Number of solutions of equation \[ 3^{2\sin^2x}+3^{2\cos^2x}=6 \] lying in interval \[ [-\pi,\pi] \] is

• A. 2
• B. 4
• C. 3
• D. 1

Hint:

Whenever exponential terms involve \(\sin^2x\) and \(\cos^2x\), use \(\sin^2x+\cos^2x=1\).

Answer 1

The Correct Option is B

Concept: Use identity \[ \sin^2x+\cos^2x=1 \] and substitution.

Step 1: Substitute variable.
Let \[ a=3^{2\sin^2x} \] Then \[ 3^{2\cos^2x} = 3^{2(1-\sin^2x)} \] \[ =\frac9a \] Equation becomes \[ a+\frac9a=6 \]

Step 2: Solve quadratic.
\[ a^2-6a+9=0 \] \[ (a-3)^2=0 \] \[ a=3 \] Thus \[ 3^{2\sin^2x}=3 \] \[ 2\sin^2x=1 \] \[ \sin^2x=\frac12 \]

Step 3: Find solutions.
\[ \sin x=\pm\frac1{\sqrt2} \] In interval \[ [-\pi,\pi] \] solutions: \[ -\frac{3\pi}{4},-\frac{\pi}{4},\frac{\pi}{4},\frac{3\pi}{4} \] Total \[ 4 \] Hence \[ \boxed{4} \]