How do you solve $cos^2 x-sin^2 x=sin x$ ; for $-pi<x<=pi$ ?

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Answer 1 · 2016-04-19T18:17:21

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

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

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

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

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

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

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

$sinx=1/2 or sinx=-1$

$x=sin^-1(1/2) or x=sin^-1 (-1)$

$x=pi/6 +2pin, (5pi)/6+2pin or (3pi)/2 +2pin$

$n=-1, x=-(11pi)/6,-(7pi)/6,-pi/2$

$n=0, x=pi/6, (5pi)/6, (3pi)/2$

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

How do you solve cos^2 x-sin^2 x=sin xcos^2 x-sin^2 x=sin x; for -pi<x<=pi-pi<x<=pi?