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Mathematics 15 Online
OpenStudy (anonymous):

Solve: sin x + sin 2x = 0

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?

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

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 :(

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 ?

OpenStudy (anonymous):

replace a and b with the values we substituted them for sinx(1+2cosx)=0

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