# sin2x + cos^2x = 0 general solution

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#1
i cant seem to get the second solution from sinx + cosx = 0

i can make it into sqrt2sin(x + pi/4) = 0

just get confused how to work the general solution out from here.
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#2
okay i will write out my working in full:

sin2x + 2cos^2x = 0

2sinxcosx + 2cos^2x

2cosx(sinx + cosx) = 0

2cosx = 0

x = 2npi +/- pi/2

sinx + cosx = 0

sqrt2sin(x + pi/4) = 0

how do i work out the generalo soution for this last line.

the book has second solution equal to npi + 3pi/4

i just cant see how to do it.

is it sumthing like x + pi/4 = npi and x + pi/4 = 2npi + pi

x = npi - pi/4

x = 2npi + 3pi/4

I just cant work it out - help please>?
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16 years ago
#3
sin(x + pi/4) = 0
x+pi/4 = npi + pi (n is an integer and this is still of the form kpi.)
x = npi + 3pi/4
Don't know why they've done it that way.
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#4
(Original post by SsEe)
sin(x + pi/4) = 0
x+pi/4 = npi + pi (n is an integer and this is still of the form kpi.)
x = npi + 3pi/4
Don't know why they've done it that way.
yes - tis strange. this book doesnt make learning this topic easy - it keeps changing its mind all the time - its has schizomath.
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