Math Practice Set 1 P1 12(e)

• (e) Tried solving it.

• Since α is difficult to type, I’ll use x instead.

Let $k,l,m\in \mathbb{Z}$.

Substituting sin3x=3sinx-4sin3x^3,

• $\sin (2x)=\sin (3x)$

The left-hand side (LHS) can be converted as:

• $\sin (2x)=\sin (2x+\pi (2k))=\sin (-2x+\pi (2k+1))$

The right-hand side (RHS) can be converted as:

• $\sin (3x)=\sin (3x+\pi (2k))=\sin (-3x+\pi (2k+1))$

By trying every combination of the converted LHS and RHS,

Pattern 1:

• $\sin (2x+\pi (2k))=\sin (3x+\pi (2l))$
• $2x+\pi (2k)=3x+\pi (2l)$
• $x=2\pi m$

Since $0\leqq x\leqq \pi$, $x=0$

Pattern 2:

• $\sin (2x+\pi (2k))=\sin (-3x+\pi (2l+1))$
• $2x+\pi (2k)=-3x+\pi (2l+1)$
• $5x=\pi (2m+1)$
• $x=\frac{\pi (2m+1)}{5}$

Since $0\leqq x\leqq \pi$, $x=\frac{\pi }{5},\frac{3\pi }{5}$

Pattern 3:

• $\sin (-2x+\pi (2k+1))=\sin (-3x+\pi (2l+1))$
• $-2x+\pi (2k+1)=-3x+\pi (2l+1)$
• $x=2\pi m$

Since $0\leqq x\leqq \pi$, $x=0$

Pattern 4:

• $\sin (-2x+\pi (2k+1))=\sin (3x+\pi (2l))$
• $-2x+\pi (2k+1)=3x+\pi (2l)$
• $x=\pi (2m+1)$

Since $0\leqq x\leqq \pi$, $x=\pi$

Therefore, $x=0,\frac{\pi }{5},\frac{3\pi }{5},\pi$

This matches the observation of the graph.