Get the answer to "Solve sin⁡(x)−sin⁡(π3−x)=32"

Judith Warner

Answered question

2022-04-22

Solve

\sin (x) - \sin (\frac{π}{3} - x) = \sqrt{\frac{3}{2}}

zavesiljid

Beginner2022-04-23Added 21 answers

\sin (x) - \sin (\frac{π}{3} - x) = \sqrt{\frac{3}{2}}

\Rightarrow \sin (x) - \cos (\frac{π}{6} + x) = \sqrt{\frac{3}{2}}

\Rightarrow \frac{3 \sin (x)}{2} - \frac{1}{2} \sqrt{3} \cos (x) = \sqrt{\frac{3}{2}}

\Rightarrow - \sqrt{3} (\frac{\cos (x)}{2} - \frac{1}{2} \sqrt{3} \sin (x)) = \sqrt{\frac{3}{2}}

\Rightarrow - \sqrt{3} (\cos (\frac{π}{3}) \cos (x) - \sin (\frac{π}{3}) \sin (x)) = \sqrt{\frac{3}{2}}

\Rightarrow - \sqrt{3} \sin (\frac{π}{6} - x) = \sqrt{\frac{3}{2}}

\sin (\frac{π}{6} - x) = - \frac{1}{\sqrt{2}}

\Rightarrow \frac{π}{6} - x = \frac{5 π}{4} + 2 π n_{1} \Rightarrow \frac{π}{6} - x = \frac{7 π}{4} + 2 π n_{2}

- x = \frac{13 π}{12} + 2 π n_{1} \Rightarrow - x = \frac{19 π}{12} + 2 π n_{2}

x = - \frac{13 π}{12} - 2 π n_{1} x = - \frac{19 π}{12} + 2 π n_{2}

With

n_{1}, n_{2} \in Z

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