How do you solve #2sin^2x-sinx=1#?

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Answer 1 · 2016-10-31T03:26:32

Add a #-sinx# to both sides to get

#2*sin^2x-sinx-sinx=1-sinx#

#2*sin^2x-2sinx=1-sinx#

#2sinx(sinx-1)-(1-sinx)=0#

#2*sinx(sinx-1)+(sinx-1)=0#

#(sinx-1)*(2*sinx+1)=0#

From the last equation we have

#sinx=1=>x=1/2 (4*pi*n+pi)# where #n# integer

and

#sinx=-1/2# with solutions

#x_1 = 1/6 (12*pi*k-pi)#

and

#x_2 = 1/6 (12*pi*k+7*pi)#

where #k# integer

Answer 2 · 2016-10-31T03:52:33

#x=((7pi)/6+2npi)#, #x=((11pi)/6+2npi)# and #x=(pi/2+2npi)# where #n in NN#.

#2sin^2x-sinx=1#

Subtract #1# from each side.#

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

Factor

#(2sinx+1)(sinx-1)=0#

Set each factor equal to zero and solve.

#2sinx+1=0# and #sinx-1=0#

#sinx=-1/2# and #sinx=1#

For #sinx=-1/2# the unit circle shows that #x=(7pi)/6# and #x=(11pi)/6# within #0<=x<2pi#

#=>x=(7pi)/6+2npi# and #x=(11pi)/6+2npi# where #n# is an integer.

For #sinx=1#, the unit circle shows that #x=pi/2# for #0<=x<2pi#

#=> x=pi/2+2npi# where #n# is an integer.