Solve on the interval [0,3π[ the equation:? Sin^2(x)=tan(x) We have t=tan(x) Please help me to solve this exercise

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Answer 1 · 2018-03-31T13:02:46

#0 , pi , 2pi, 3pi#

#sin^2(x)=tan(x)#

Identity:

#color(red)bb(tan(x)=sin(x)/cos(x)#

#sin^2(x)-sin(x)/cos(x)=0#

Add:

#(cos(x)sin^2(x)-sin(x))/cos(x)=0#

#(cos(x)sin^2(x)-sin(x))=0#

Factor:

#sin(x)(cos(x)sin(x)-1)=0#

#sin(x)=0=>x=0 , pi , 2pi, 3pi#

#cos(x)sin(x)-1=0#

#cos(x)sin(x)=1#

Identity:

#color(red)bb(sin(x)cos(x)=1/2sin(2x)#

#1/2sin(2x)=1#

#sin(2x)=2#

No solutions for this.

#-1<=sin(x)<=1#