Here it is given that, cos4x=cos2x
Which implies,
cos4x-cos2x=0
Now we know that,
cosA-cosB=-2sin\( ( \dfrac{A+B}{2})sin( \dfrac{A-B}{2})\)
Therefore we have,
\(-2sin= ( \dfrac{4x+2x}{2})sin( \dfrac{4x-2x}{2})=0\)
sin3xsinx=0
Hence we have, sin3x=0
Or, sinx=0
Therefore, 3x=nπ
or x=nπ ,where \( n\in Z\)
\(x= \dfrac{n\pi}{3}\)
or x=nπ ,where \( n\in Z\)
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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