Answered: "Finding a for sin⁡(4a+π6)=sin⁡(2a+π5)"

STATEMENT2z4

Answered question

2022-03-01

Finding a for

\sin (4 a + \frac{π}{6}) = \sin (2 a + \frac{π}{5})

Answer & Explanation

emeriinb4r

Beginner2022-03-02Added 10 answers

$\sin (4 a + \frac{π}{6}) = \sin (2 a + \frac{π}{5})$
$\sin (4 a + \frac{π}{6}) - \sin (2 a + \frac{π}{5}) = 0$
Consider the formula: $\sin a - \sin b = 2 \cos \frac{a + b}{2} \sin \frac{a - b}{2}$
and $A \cdot B = 0 \Rightarrow A = 0 \lor B = 0$
$2 \cos \frac{6 a + \frac{11 π}{30}}{2} \sin \frac{2 a - \frac{π}{30}}{2} = 0$
$\cos \frac{6 a + \frac{11 π}{30}}{2} = 0 \lor \sin \frac{2 a - \frac{π}{30}}{2} = 0$
$\frac{6 a + \frac{11 π}{30}}{2} = 0 \lor \cos m = 0 \Rightarrow m = \frac{π}{2} + k π$
Hence we have:
$\frac{6 a + \frac{11 π}{30}}{2} = \frac{π}{2} + k π \Rightarrow a = \frac{k π}{3} + \frac{19 π}{180}, k \in Z$
$\frac{2 a - \frac{π}{30}}{2} = n \Rightarrow \sin n = 0 \Rightarrow n = k π$
Hence we have:
$\frac{2 a - \frac{π}{30}}{2} = k π \Rightarrow a = k π + 3, k \in Z$

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