Animated Solution for Mathematics - Trigonometry: The number of solutions of $\sin 3x=\cos 2x$, in the interval $\left[\frac{\pi }{2},\pi \right]$ is :-

We are tasked with finding the number of solutions for the equation $\sin 3x=\cos 2x$ within the closed interval $x\in [\frac{\pi }{2},\pi ]$. This problem requires careful handling of trigonometric periodicity and strict adherence to the specified boundary conditions.

The primary obstacle is the mismatch between the sine and cosine functions. To resolve this, we invoke the complementary angle identity: $\cos \theta =\sin (\frac{\pi }{2}-\theta )$.

Applying this identity, our equation transforms into:

The general solution for the equation $\sin A=\sin B$ is given by $A=n\pi +(-1{)}^{n}B$, where $n\in \mathbb{Z}$. This formula accounts for the periodic nature of the sine function.

Substituting our specific angles into the general solution, we obtain:

Even Case

We split the analysis based on the parity of $n$. For the Even Case, let $n=2k$ (where $k\in \mathbb{Z}$). Here, $(-1{)}^{2k}=1$, and the equation becomes:

Rearranging for $x$, we get $5x=2k\pi +\frac{\pi }{2}$, which simplifies to:

We now test values of $k$ to find solutions within $[\frac{\pi }{2},\pi ]$:

For $k=1$: $x=\frac{5\pi }{10}=\frac{\pi }{2}$. This is a valid solution as it lies on the boundary. For $k=2$: $x=\frac{9\pi }{10}$. Since $\frac{9\pi }{10}<\pi$, this is a valid solution. * For $k=3$: $x=\frac{13\pi }{10}>\pi$. This exceeds our interval.

Odd Case

For the Odd Case, let $n=2k+1$. Here, $(-1{)}^{2k+1}=-1$, and the equation becomes:

Expanding and simplifying the expression:

Testing values of $k$:

For $k=0$: $x=\frac{\pi }{2}$. This is the same solution found previously. For $k=1$: $x=\frac{5\pi }{2}$, which is clearly outside our interval.

By methodically exploring both branches and checking the boundaries, we have identified the distinct solutions as $x=\frac{\pi }{2}$ and $x=\frac{9\pi }{10}$.

There are exactly 2 distinct solutions in the interval $[\frac{\pi }{2},\pi ]$.