Solve #2tan^2xsinx-tan^2x=0#, if #x in[0^@,360^@]#?

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Answer 1 · 2018-01-13T03:40:57

#x in(0^o,30^o,150^o,180^o,360^o)#

#2tan^2xsinx-tan^2x=0#

#2sin^2x/cos^2xsinx-sin^2x/cos^2x=0#

Multiply through by #cos^2x#:

#2sin^2xsinx-sin^2x=0#

#2sin^3x-sin^2x=0#

#sin^2x(2sinx-1)=0#

So either:

#sin^2x=0# or #2sinx-1=0#

Solving for #sin^2x=0#

#-> sin(x)=0-> x=sin^(-1)(0)#

In the interval of #0^o# to #360^o# 3 possible values satisfy this, namely: #0^o,180^o# and #360^o#.

Solving:

#2sinx-1=0-> x=sin^(-1)(1/2)#

To values satisfy this in the interval, namely:

#x=30^o# and #x=180^o-30^o=150^o#

So #x in(0^o,30^o,150^o,180^o,360^o)#