find-all-the-values-of-in-the-interval-0-2pi-for-which-sin-3-sin-3cos-2-

Question Number 97994 by hardylanes last updated on 10/Jun/20

f i n d a l l t h e v a l u e s o f θ, i n t h e i n t e r v a l 0 ⩽ θ ⩽ 2 π

f o r w h i c h \sin 3 θ - \sin θ = \sqrt{3 \cos 2 θ}

Answered by Rio Michael last updated on 10/Jun/20

i pressume your question is

\sin 3 θ - \sin θ = \sqrt{3} \cos 2 θ for 0 ⩽ θ ⩽ 2 π

R ecall \sin A - \sin B = 2 \cos (\frac{A + B}{2}) \sin (\frac{A - B}{2})

\Rightarrow 2 \cos (\frac{3 θ + θ}{2}) \sin (\frac{3 θ - θ}{2}) = \sqrt{3} \cos 2 θ

\Rightarrow 2 \cos 2 θ \sin θ = \sqrt{3} \cos 2 θ

\Rightarrow 2 \cos 2 θ \sin θ - \sqrt{3} \cos 2 θ = 0

\cos 2 θ (2 \sin θ - \sqrt{3}) = 0

\Leftrightarrow \cos 2 θ = 0 \Rightarrow 2 θ = 2 π n \pm \frac{π}{2} \Rightarrow θ = \frac{1}{2} (2 π n \pm \frac{π}{2}) n \in Z^{+}

\sin θ = \frac{\sqrt{3}}{2} \Rightarrow π n + {(- 1)}^{n} \frac{π}{3} n \in Z^{+}

general solution θ = {\begin{cases} \frac{1}{2} (2 π n \pm \frac{π}{2}) \\ π n + {(- 1)}^{n} \frac{π}{3} \end{cases} n \in Z^{+}

n = 0, \Rightarrow θ = \frac{π}{4}, \frac{π}{3}

n = 1, \Rightarrow θ = \frac{4 π}{3}, \frac{3 π}{4}

\Rightarrow solution S = {\frac{π}{4}, \frac{π}{3}, \frac{4 π}{3}, \frac{3 π}{4}}

Commented by hardylanes last updated on 10/Jun/20

thanks☆

Commented by Rio Michael last updated on 10/Jun/20

you are welcome .