How do you solve #cos^2(x)+sin=1#?

Question

23263 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-29T06:21:05

Possible solution within the domain #[0,2pi]# are #{0, pi/2, pi, 2pi}#

#cos^2(x)+sinx=1# can be written as #sinx=1-cos^2x=sin^2x#

(I have assumed that by #cos^2(x)+sin=1#, one meant #cos^2(x)+sinx=1#

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

#sinx(sinx-1)=0#

Hence either #sinx=0# or #sinx=1#

Hence, possible solution within the domain #[0,2pi]# are

#{0, pi/2, pi, 2pi}#

Answer 2 · 2016-02-29T06:45:28

Under limit [0,2#pi#], #x=0,pi,2pi# or #x=pi/2#

given that

#cos^2x + sin x =1#

#=>sinx=1-cos^2x#

#=>sinx=sin^2x#

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

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

#=>sinx=0 (or) sinx-1=0#

IF Sin(x) =0 :-

Then #x=0,pi,2pi,3pi......#

IF sin(x)=1 :-

Then #x=pi/2,(5pi)/2......#