Here it is given that,
secx=2
Now we know that
\( sec= \dfrac{\pi}{3}=2\) and
\( sec\dfrac{5\pi}{3}=sec(2\pi- \dfrac{\pi}{3})\)
\( sec\dfrac{\pi}{3}=2\)
Therefore, the principal solutions are
\( \dfrac{\pi}{3}\) and \( \dfrac{5\pi}{3}\)
Now, \( secx=sec \dfrac{\pi}{3}\)
and we know ,
\( secx=\dfrac{1}{cosx}\)
Therefore , we have,
\( cosx=cos \dfrac{\pi}{3}\)
Which implies,
\( x=2nπ\pm \dfrac{\pi}{3},n\in Z\)
Answered by Abhisek | 1 year agoprove that \(sin \dfrac{8π}{3} cos \dfrac{23π}{6} + cos \dfrac{13π}{3} sin \dfrac{35π}{6} = \dfrac{1}{2}\)
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