How do you solve ${cos}^{2} x - {sin}^{2} x = sin x$ $cos^2 x-sin^2 x=sin x$ ; for $- π < x \leq π$ $-pi<x<=pi$ ?

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Answer 1 · 2016-04-19T18:17:21

$S = {- \frac{π}{2}, \frac{π}{6}, \frac{5 π}{6}}$ $S={-pi/2, pi/6,(5pi)/6}$

Use Property: ${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x+sin^2x=1$

$1 - {sin}^{2} x - {sin}^{2} x = sin x$ $1-sin^2x-sin^2x=sinx$

$1 - 2 {sin}^{2} x - sin x = 0$ $1-2sin^2x-sinx=0$

$2 {sin}^{2} x + sin x - 1 = 0$ $2sin^2x+sinx-1=0$

$(2 sin x - 1) (sin x + 1) = 0$ $(2sinx-1)(sinx+1)=0$

$2 sin x - 1 = 0 or sin x + 1 = 0$ $2sinx-1=0 or sinx+1=0$

$sin x = \frac{1}{2} or sin x = - 1$ $sinx=1/2 or sinx=-1$

$x = {sin}^{- 1} (\frac{1}{2}) or x = {sin}^{- 1} (- 1)$ $x=sin^-1(1/2) or x=sin^-1 (-1)$

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n or \frac{3 π}{2} + 2 π n$ $x=pi/6 +2pin, (5pi)/6+2pin or (3pi)/2 +2pin$

$n = - 1, x = - \frac{11 π}{6}, - \frac{7 π}{6}, - \frac{π}{2}$ $n=-1, x=-(11pi)/6,-(7pi)/6,-pi/2$

$n = 0, x = \frac{π}{6}, \frac{5 π}{6}, \frac{3 π}{2}$ $n=0, x=pi/6, (5pi)/6, (3pi)/2$

$S = {- \frac{π}{2}, \frac{π}{6}, \frac{5 π}{6}}$ $S={-pi/2, pi/6,(5pi)/6}$

How do you solve cos^2 x-sin^2 x=sin xcos2x−sin2x=sinxcos^2 x-sin^2 x=sin x; for -pi<x<=pi−π<x≤π-pi<x<=pi?