Question

The number of solutions of \( 5^{1+|\sin x|+|\sin x|^2+|\sin x|^3+\cdots} = 25 \) for \( x \in (-\pi, \pi) \) is:

• A. 2
• B. 0
• C. 4
• D. infinite

Hint:

Infinite geometric series converge only when the absolute value of the common ratio is less than 1.

Answer 1

The Correct Option is C

Step 1: Recognize the infinite geometric series in the exponent.
The exponent is \( 1 + |\sin x| + |\sin x|^2 + |\sin x|^3 + \cdots \). This is an infinite geometric series with the first term \( a = 1 \) and the common ratio \( r = |\sin x| \). 
Step 2: Determine the condition for the convergence of the geometric series.
For an infinite geometric series to converge, \( |r|<1 \), so \( ||\sin x||<1 \), which means \( |\sin x|<1 \). 
Step 3: Find the sum of the infinite geometric series.
The sum is \( S = \frac{a}{1 - r} = \frac{1}{1 - |\sin x|} \). 
Step 4: Substitute the sum back into the original equation. \[ 5^{\frac{1}{1 - |\sin x|}} = 25 \] 
Step 5: Solve for \( |\sin x| \). \[ 5^{\frac{1}{1 - |\sin x|}} = 5^2 \implies \frac{1}{1 - |\sin x|} = 2 \implies 1 = 2 - 2|\sin x| \implies |\sin x| = \frac{1}{2} \] 
Step 6: Find the values of \( x \) in the interval \( (-\pi, \pi) \) that satisfy \( |\sin x| = \frac{1}{2} \).
For \( \sin x = \frac{1}{2} \), \( x = \frac{\pi}{6}, \frac{5\pi}{6} \). For \( \sin x = -\frac{1}{2} \), \( x = -\frac{\pi}{6}, -\frac{5\pi}{6} \). 
Step 7: Count the number of solutions.
There are 4 distinct solutions.