find the general solution of 2sin²x + sin²2x = 2
ssara:
question is from which book?
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Answered by
6
it's very easy
2sin^2x+ (2sinx.cosx)^2=2
2sin^2x+4sin^2x.cos^2x=2
sin^2x+2sin^2x.cos^2x=1
2sin^2x.cos^2x= 1-sin^2x
2sin^2x.cos^2x= cos^2x
2sin^2x=1
sin^2x= 1/2
sinx= 1/√2
=> x= pi/4 or 45 degrees
Since the value of sinx repeats after every 180 deg turn, so....you would get the same value of sinx if you add pi to pi/4 or 2pi to pi/4. The solutions are countless.
x= pi/4, 3pi/4 etc
2sin^2x+ (2sinx.cosx)^2=2
2sin^2x+4sin^2x.cos^2x=2
sin^2x+2sin^2x.cos^2x=1
2sin^2x.cos^2x= 1-sin^2x
2sin^2x.cos^2x= cos^2x
2sin^2x=1
sin^2x= 1/2
sinx= 1/√2
=> x= pi/4 or 45 degrees
Since the value of sinx repeats after every 180 deg turn, so....you would get the same value of sinx if you add pi to pi/4 or 2pi to pi/4. The solutions are countless.
x= pi/4, 3pi/4 etc
thank bhai
mention Not
how did you multiply cos x in the first step?
konsa formula use kiya
wait
you are wrong brother.........
you have copied the answer from yahoo answers:)))
sin2x = 2sinxcosx
Answered by
4
the answer is 45 degree.
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