Questions
General solution of the equation is
a
x=(4n+1)π4
b
x=(2n+1)π4
c
x=(6n+1)π6
d
x=(3n+1)π3
detailed solution
Correct option is A
We have sin4x+cos4x=sinxcosx⇒1−2sin2xcos2x=sinxcosx⇒1−2t2=t where t=sinxcosx⇒2t2+t−1=0⇒2t2+2t−t−1=0⇒2t(t+1)−1(t+1)=0⇒(2t−1)(t+1)=0⇒t=12or t=−1⇒sinxcosx=12 or sinxcosx=−1⇒sin2x=1 or sin2x=−2not possible⇒2x=2nπ+π2⇒x=nπ+π4,n∈z =(4n+1)π4Talk to our academic expert!
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