How do you find the general solutions for #sin^2x - 2sinx = 1#?

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Answer 1 · 2018-03-06T11:38:20

#pi-arcsin(1-sqrt(2))+npi#

#4pin+arcsin(1-sqrt(2))#

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

This is just a quadratic in #sinx#:

Let #u=sinx#

Then:

#u^2-2u-1=0#

Using the quadratic formula:

#(-(-2)+-sqrt((-2)^2-4(1)(-1)))/(2(1)#

#(2+-sqrt(4+4))/(2)#

#(2+-sqrt(8))/(2)#

#(2+-2sqrt(2))/(2) = 1+sqrt(2) , color(white)(888)1-sqrt(2)#

#u=sinx#

#:.#

#sinx=1+sqrt(2)#, #color(white)(888)sinx=1-sqrt(2)#

#x=arcsin(sinx)=arcsin(1+sqrt(2))# ( no real solutions )*

#x=arcsin(sinx)=arcsin(1-sqrt(2))=arcsin(1-sqrt(2))#

#2pi + arcsin(1-sqrt(2))color(white)(88)# IV quadrant

#pi-arcsin(1-sqrt(2))color(white)(888)# III quadrant

These values are in the interval:

#0<=x<=2pi#

For the general solution:

#pi-arcsin(1-sqrt(2))+npi#

#4pin+arcsin(1-sqrt(2))#

For integers #bbn#

*

domain of #arcsin(x)#

#[-1,1]#

#1+sqrt(2)~~2.414213562#