Tinkutara Equation Editor: Math Forum Question #: 103823


Question Number 103823 by bobhans last updated on 17/Jul/20

	Answered by Dwaipayan Shikari last updated on 17/Jul/20
	$((sinx)/(cosx))=cos2x(4−(1/(sin2x))) ((sinx)/(cosx))=((2sin4x−cos2x)/(2sinxcosx)) 2sin^2 x=2sin4x−cos2x 1−cos2x=2sin4x−cos2x sin4x=(1/2) 4x=2kπ+(−1)^k (π/6) x=((kπ)/2)±(π/(24))=(12k±1)(π/(24))= { k,m∈Z (general solution$
	Answered by bobhans last updated on 17/Jul/20