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Discussion Response Post to: trigo

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puneet (2841)

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There seems to be a flaw in your approach ....

-2sinx.cos3x = 0

then sinx = 0

cos 3x = 0

so,

sinx = 0

x= n

But note the difference here .......

cos 3x = 0

that 3x lies on the y axis ....

So 3x= (2n+1)

or x = (2n+1)

so finally solving like this we get x = n

or x = (2n+1)

Now take the second approach ...

sin 2x(1- 2cos2x) = 0

then sin2x= 0

and cos 2x = 1/2

sin2x= 0

2x= n

or x = n

cos 2x = 1/2

2x = 2n

/3 or 2x = 2n

So x = n

/6 or n

Thus this yields solution x = n

/2 , n

/6 or n

Now one can easily establish a one to one correspondance between two solutions .. both soultions represent same set ...

cheers

Puneet Agrawal IIT Delhi

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