Asked by Anonymous
how do i solve
sin(x)+2cos(2x)=0
sin(x)+2cos(2x)=0
Answers
Answered by
MathMate
Use the double angle identity to express cos(2x) in terms of sin(x)
cos(2x)
=cos^2(x)-sin^2(x)
=1-2sin^2(x)
sin(x)-2(1-sin^2(x))=0
2sin^2(x)+sin(x)-2=0
(2sin(x)-1)(sin(x)+1)=0
=>
sin(x)=1/2 or
sin(x)=-1
Can you take it form here?
cos(2x)
=cos^2(x)-sin^2(x)
=1-2sin^2(x)
sin(x)-2(1-sin^2(x))=0
2sin^2(x)+sin(x)-2=0
(2sin(x)-1)(sin(x)+1)=0
=>
sin(x)=1/2 or
sin(x)=-1
Can you take it form here?
Answered by
MathMate
I take it that you're OK with it.
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