# Trigonometric Equation involving Double Angle and Compound Angle Formula

Solve $\sin x + \sin 2x + \sin 3x = 0$ for $0 \le x \le 2 \pi$.

## Part 1

Show that $\sin 3x = 3 \sin x-4 \sin^3 x$.

\begin{align} \sin 3x &= \sin (x + 2x) \\ &= \sin x \cos 2x + \cos x \sin 2x &\color{green}{\text{Compound Angle Formula}} \\ &= \sin x (1-2 \sin^2 x) + \cos x (2 \sin x \cos x) &\color{green}{\text{Double Angle Formula}} \\ &= \sin x-2 \sin^3 x + 2 \sin x \cos^2 x \\ &= \sin x-2 \sin^3 x + 2 \sin x (1-\sin^2 x) &\color{green}{\sin^2 x + \cos^2 x = 1} \\ &= \sin x-2 \sin^3 x + 2 \sin x-2 \sin^3 x \\ &= 3 \sin x-4 \sin^3 x \end{align}

## Part 2

Show that $\sin x + \sin 3x = 2 \sin 2x \cos x$.

\begin{align} \sin x + \sin 3x &= \sin x + 3 \sin x-4 \sin^3 x &\color{green}{\text{Result of Part 1}} \\ &= 4 \sin x-4 \sin^3 x \\ &= 4 \sin x (1-\sin^2 x) \\ &= 4 \sin x \cos^2 x &\color{green}{\sin^2 x + \cos^2 x = 1} \\ &=2 \times 2 \sin x \cos x \times \cos x \\ &= 2 \sin 2x \cos x &\color{green}{\text{Compound Angle Formula}} \end{align}

## Part 3

Solve $\sin x + \sin 2x + \sin 3x = 0$ for $0 \le x \le 2 \pi$.

\begin{align} \sin x + \sin 2x + \sin 3x &= 0 \\ (\sin x + \sin 3x) +\sin 2x &=0 \\ 2 \sin 2x \cos x + \sin 2x &= 0 &\color{green}{\text{Result of Part 2}} \\ \sin 2x (2 \cos x + 1) &= 0 \\ \sin 2x &= 0 \text{ or } 2 \cos x + 1 = 0 \\ 2x &= 0, \pi, 2 \pi, 3 \pi, 4 \pi, 5 \pi, \cdots &\color{green}{\text{for } \sin 2x =0} \\ x &= 0, \displaystyle \frac{\pi}{2}, \pi, \frac{3 \pi}{2}, 2 \pi, \frac{5 \pi}{2}, \cdots \\ &= 0, \frac{\pi}{2}, \pi, \frac{3 \pi}{2}, 2 \pi &\color{green}{\text{for } 0 \le x \le 2 \pi} \\ x &= \pi-\frac{\pi}{3}, \pi + \frac{\pi}{3}, 3 \pi – \frac{\pi}{3}, 3 \pi + \frac{\pi}{3}, \cdots &\color{green}{\text{for } \cos x = -\frac{1}{2} } \\ &= \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{8 \pi}{3}, \frac{10 \pi}{3}, \cdots \\ &= \frac{2 \pi}{3}, \frac{4 \pi}{3} &\color{green}{\text{for } 0 \le x \le 2 \pi} \\ \require{AMSsymbols} \therefore x &= 0, \frac{\pi}{2}, \pi, \frac{3 \pi}{2}, \frac{3 \pi}{2}, \frac{4 \pi}{3}, 2 \pi \end{align}

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