How do you solve $cos^2 x-sin^2 x=sin x$ over the interval $-pi<x<=pi$ ?

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Answer 1 · 2015-04-16T00:59:33

Write: $cos^2(x)=1-sin^2(x)$ so that you get:

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

Set: $sin(x)=t$ and get:

$-2t^2-t+1=0$
$t_1=-1$
$t_2=1/2$

but $sin(x)=t$

And in your interval:

$sin(x)=-1$ and $x=-pi/2$
$sin(x)=1/2$ and $x=pi/6 and x=5pi/6$

![http://www.hyper-ad.com/tutoring/math/trig/The%20Sine%20Function.html](useruploads.socratic.org)

Hope it helps.

How do you solve cos^2 x-sin^2 x=sin xcos^2 x-sin^2 x=sin x over the interval -pi<x<=pi-pi<x<=pi?