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

Question

23918 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 $x=pi/2,(5pi)/2......$

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