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Mathematics 19 Online
OpenStudy (anonymous):

sin2x = cosx

OpenStudy (anonymous):

sin2x = 2sinxcosx helps ?

OpenStudy (anonymous):

so you have 2sinxcosx=cosx 2sinx=1 i guess you know how to solve it now.

OpenStudy (anonymous):

i am realy struglling plzz help me :(

OpenStudy (anonymous):

sinx=1/2

OpenStudy (anonymous):

and we know sin(30) = 1/2 so you have sinx = sin30 so the solutions are: x = 30 + 360k x = 180-30+360k = 150+360k

OpenStudy (anonymous):

tnxxx u soo much

OpenStudy (anonymous):

than its says solve for x if e [-360;360]

OpenStudy (anonymous):

@Coolsector, dividing both sides by \(\cos x\) removes a solution. \[2\sin x\cos x=\cosx\\ 2\sin x\cos x -\cos x=0\\ \cos x(2\sin x-1)=0\]

OpenStudy (anonymous):

@SithsAndGiggles is right. sorry for that.

OpenStudy (anonymous):

so we need to add the solution for cosx = 0

OpenStudy (anonymous):

so do you know how to solve cosx=0 ?

OpenStudy (anonymous):

explain in detail i think i am getting it???

OpenStudy (anonymous):

cosx(2sinx-1) = 0 gives two options 2sinx-1 = 0 which we found the solution for it and cosx=0 the solution for it is : x=90+180k

OpenStudy (anonymous):

so we have: x=90+180k x = 30 + 360k x = 180-30+360k = 150+360k were k can be 0,-+1,-+2...

OpenStudy (anonymous):

OpenStudy (anonymous):

help me solve this?

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