Solving Trigonometric Equation in Interval

We need to find the number of solutions for the equation $sin(3x) = cos(2x)$ in the interval $\left[ \frac{\pi}{2}, \pi \right]$.

Equation Transformation

First, rewrite the equation using trigonometric identities. We know that $cos(\theta) = sin(\frac{\pi}{2} - \theta)$. Applying this:

$sin(3x) = sin\left(\frac{\pi}{2} - 2x\right)$

General Solution

The general solution for $sin(A) = sin(B)$ is given by $A = k\pi + (-1)^k B$, where $k$ is an integer.

In our case, $A = 3x$ and $B = \frac{\pi}{2} - 2x$. So, we have:

$3x = k\pi + (-1)^k \left(\frac{\pi}{2} - 2x\right)$

We consider two cases for $k$:

Case 1: $k$ is even ($k = 2n$, where $n$ is an integer)

$3x = 2n\pi + (-1)^{2n} \left(\frac{\pi}{2} - 2x\right)$

$3x = 2n\pi + 1 \cdot \left(\frac{\pi}{2} - 2x\right)$

$3x = 2n\pi + \frac{\pi}{2} - 2x$

$5x = 2n\pi + \frac{\pi}{2}$

$x = \frac{2n\pi}{5} + \frac{\pi}{10} = \frac{(4n+1)\pi}{10}$

Case 2: $k$ is odd ($k = 2n+1$, where $n$ is an integer)

$3x = (2n+1)\pi + (-1)^{2n+1} \left(\frac{\pi}{2} - 2x\right)$

$3x = (2n+1)\pi - 1 \cdot \left(\frac{\pi}{2} - 2x\right)$

$3x = (2n+1)\pi - \frac{\pi}{2} + 2x$

$x = (2n+1)\pi - \frac{\pi}{2}$

$x = 2n\pi + \pi - \frac{\pi}{2} = 2n\pi + \frac{\pi}{2}$

Finding Solutions in the Interval $\left[ \frac{\pi}{2}, \pi \right]$

Now, we find the values of $n$ for which $x$ lies in the interval $\left[ \frac{\pi}{2}, \pi \right]$ (i.e., $0.5\pi \le x \le 1\pi$).

Solutions from Case 1 ($x = \frac{(4n+1)\pi}{10}$)

• For $n=0$, $x = \frac{(4(0)+1)\pi}{10} = \frac{\pi}{10}$. (Not in interval)
• For $n=1$, $x = \frac{(4(1)+1)\pi}{10} = \frac{5\pi}{10} = \frac{\pi}{2}$. (In interval)
• For $n=2$, $x = \frac{(4(2)+1)\pi}{10} = \frac{9\pi}{10}$. (In interval)
• For $n=3$, $x = \frac{(4(3)+1)\pi}{10} = \frac{13\pi}{10}$. (Not in interval)

Solutions from Case 2 ($x = 2n\pi + \frac{\pi}{2}$)

• For $n=0$, $x = 2(0)\pi + \frac{\pi}{2} = \frac{\pi}{2}$. (In interval, already found)
• For $n=1$, $x = 2(1)\pi + \frac{\pi}{2} = \frac{5\pi}{2}$. (Not in interval)

Counting the Solutions

The distinct solutions within the interval $\left[ \frac{\pi}{2}, \pi \right]$ are $x = \frac{\pi}{2}$ and $x = \frac{9\pi}{10}$.

Therefore, there are 2 solutions in the given interval.