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joyfrancis (1504)

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solve:
 
sin2x - 12(sinx-cosx) + 12 = 0

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saarika (172)

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sin2x =12(sinx-cosx-1)
 
2sinx cosx=12(sinx-(cosx+1))
 
4sinx/2 .cosx/2cosx =12( 2sinx/2.cosx/2 -2cos2x/2)
 
cosx/2(sinx/2.cosx)=cosx/2(6sinx/2-cosx/2)
 
therefore cosx/2=0 implies x=npi is solution  where n=(2k+1)pi
 
i am unable to proceed further  .......
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vasanth (2315)

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hey cammon
it cant be x=npi
x/2 = pi/2
x=pi
 
generalising x=2npi +- pi
 
i'll try 2 post the solution soon
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rajat (284)

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i am getting
x + (pie)/4 = 2n(pie)  3(pie)/4 .
now i do not know how to generalise this .plzzz. someone tell me how to  proceed after reaching this step .

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