How do you solve #tantheta = sqrt(2)sintheta#?

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Answer 1 · 2017-03-31T00:08:21

#theta = (pm k pi uu pm pi/4 + 2kpi)# for #k in ZZ#

#tan(theta)=sqrt(2)sin(theta)# or

#tan(theta)(1-sqrt(2)cos(theta))=0#

#{(tan(theta)=0->theta=pm k pi),(cos(theta)=1/sqrt(2)->theta = pm pi/4 + 2kpi):}#

#theta = (pm k pi uu pm pi/4 + 2kpi)# for #k in ZZ#

Answer 2 · 2017-03-31T02:00:36

#theta={pi/4, (7pi)/4}#

Here's an alternative approach. Note that #tantheta = sin theta/costheta#.

#sintheta/costheta = sqrt(2)sintheta#

Multiply both sides by #costheta#.

#sintheta/costheta * costheta = sqrt(2)sintheta*costheta#

Now recognize that #sin2theta = 2sinthetacostheta#.

#sintheta = sqrt(2)sinthetacostheta#

Multiply the right side by #2/2#
#sintheta = 2sqrt(1/2)sinthetacostheta#

#sintheta = sqrt(1/2)sin2theta#

#sintheta/(2sinthetacostheta) = 1/sqrt(2)#

#1/(2costheta) = 1/sqrt(2)#

#sqrt(2) = 2costheta#

#sqrt(2)/2 = costheta#

This is the rationalized form of #1/sqrt(2)#.

#1/sqrt(2) = costheta#

This has solutions #theta = pi/4, (7pi)/4#.

Hopefully this helps!