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

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Answer 1 · 2016-02-29T06:21:05

Possible solution within the domain $[0, 2 π]$ $[0,2pi]$ are ${0, \frac{π}{2}, π, 2 π}$ ${0, pi/2, pi, 2pi}$

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

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

or ${sin}^{2} x - sin x = 0$ $sin^2x-sinx=0$ or

$sin x (sin x - 1) = 0$ $sinx(sinx-1)=0$

Hence either $sin x = 0$ $sinx=0$ or $sin x = 1$ $sinx=1$

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

${0, \frac{π}{2}, π, 2 π}$ ${0, pi/2, pi, 2pi}$

Answer 2 · 2016-02-29T06:45:28

Under limit [0,2 $π$ $pi$ ], $x = 0, π, 2 π$ $x=0,pi,2pi$ or $x = \frac{π}{2}$ $x=pi/2$

given that

${cos}^{2} x + sin x = 1$ $cos^2x + sin x =1$

$\Rightarrow sin x = 1 - {cos}^{2} x$ $=>sinx=1-cos^2x$

$\Rightarrow sin x = {sin}^{2} x$ $=>sinx=sin^2x$

$\Rightarrow {sin}^{2} x - sin x = 0$ $=> sin^2x-sinx=0$

$\Rightarrow sin x (sin x - 1) = 0$ $=>sinx(sinx-1)=0$

$\Rightarrow sin x = 0 (or) sin x - 1 = 0$ $=>sinx=0 (or) sinx-1=0$

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=1cos2(x)+sin=1cos^2(x)+sin=1?