How do you determine all solutions for the equation #2sin^2theta=1-sintheta# in the domain #0^circ<=theta<360^circ#?

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Answer 1 · 2016-11-27T10:38:33

The solutions are # S={270º,30º,150º} #

Let's rewrite the equation

#2sin^2theta+sin theta-1=0#

This s a quadratic equation, #ax^2+bc+c=0#

We calculate the discriminant,

#Delta=b^2-4ac=1-4*2*-1=9#

#Delta >0#, so we have 2 real roots

#x=(-b+-sqrt(Delta))/(2a)#

#sintheta=(-1+-sqrt9)/4=(-1+-3)/4#

#sintheta=-1# or #sintheta=1/2#

We are looking for the solutions, #theta in [0, 360[#

The solutions are #(270º, 30º,150º)#

Answer 2 · 2016-11-27T10:51:00

#theta=pi/6#, #theta=5/6pi# and #theta=3/2pi#
for #0< theta <2pi#

rewrite the equation in an equivalent form

#2sin^2theta+sintheta-1=0#

this is a second order equation in the unknown #sintheta#

We can solve it, obtaining

#sintheta=(-1+-root2(1+8))/4=(-1+-3)/4#.
There are two possible solutions

#sintheta_1=-1# that means #theta_1=3/2pi+ 2 kpi#

or
#sintheta_2=1/2# that means #theta_2=pi/6+2kpi# or #theta_2=pi-pi/6+2kpi#

In total in # 0< theta<2pi # we have three solutions
#theta=pi/6#, #theta=5/6pi# and #theta=3/2pi#