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Mathematics 8 Online
OpenStudy (anonymous):

Find all solutions to the equation in the interval [0, 2π). cos x = sin (2x)

OpenStudy (anonymous):

pi/2, 3pi/2 pi/6, pi/2, 5pi/6, 3pi/2 0, pi 0, pi/6, 5pi/6, pi

OpenStudy (anonymous):

Those are the options

OpenStudy (anonymous):

start with \[\cos(x)=2\sin(x)\cos(x)\] and solve that one

OpenStudy (shamim):

ok i m trying

OpenStudy (shamim):

\[\cos x=\sin 2x\]

OpenStudy (shamim):

\[\cos x =2\sin x \cos x\]

OpenStudy (shamim):

\[cosx(1-2sinx)=0\]

OpenStudy (shamim):

\[cosx=0=\cos \frac{ \pi }{ 2 }=\cos \frac{ 3\pi }{ 2 }\]

OpenStudy (shamim):

so\[x=\frac{ \pi }{ 2 },\frac{ 3\pi }{ 2 }\]

OpenStudy (shamim):

do the same for\[1-sinx=0\]

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

except it is \(1-2\sin(x)=0\)

OpenStudy (anonymous):

making \(\sin(x)=\frac{1}{2}\) and go from there

OpenStudy (shamim):

ya

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