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OpenStudy (anonymous):
sin2x = cosx
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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
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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
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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):
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OpenStudy (anonymous):
help me solve this?
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