Mathematics
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OpenStudy (anonymous):
Solve: sin x + sin 2x = 0
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OpenStudy (anonymous):
break down sin2x into 2sinxcosx
sinx+2sinxcosx=0
sinx(1+2cosx)=0
sinx=0 or 1+2cosx=0
sinx=0 or cosx=-1/2
OpenStudy (anonymous):
where did you get the (1+2cosx) from ?
OpenStudy (cwrw238):
sinx + 2 sinx cos x = 0
sin x ( 1 + 2 cos x) = 0
sinx = 0 , cos x = -1/2
for x between 0 and 360
x = 0, 180, 360, 120, 240 degrees
OpenStudy (anonymous):
i factored out a sinx from each term on the left
OpenStudy (anonymous):
can you please explain it more?
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OpenStudy (anonymous):
sin(2x)=2sin(x)*cos(x)
OpenStudy (anonymous):
Rewrite the equation with this in mind
OpenStudy (anonymous):
sin(x)+2sin(x)*cos(x)=0
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
Now you can factor out a sin(x) from each term
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OpenStudy (anonymous):
sin(x)[1+2cos(x)]=0
OpenStudy (anonymous):
if you were to distribute the sin(x), you'd get back to
sin(x)+2sin(x)cos(x)=0
OpenStudy (anonymous):
that's just explaining how it was factored, don't distribute it
OpenStudy (anonymous):
understand that?
OpenStudy (anonymous):
hmm it is kinda confusing, I'm still trying to understand, sorry I'm kinda slow :(
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OpenStudy (anonymous):
suppose we let sinx=a and cosx=b
OpenStudy (anonymous):
a+2ab=0
OpenStudy (anonymous):
i get the part sin x + 2sinxcosx = 0 it's in the book
OpenStudy (anonymous):
a(1+2b)=0
OpenStudy (anonymous):
but i dont get the part (1+2cos x) from ?
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OpenStudy (anonymous):
replace a and b with the values we substituted them for
sinx(1+2cosx)=0