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

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

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Answer 1 · 2015-04-16T00:59:33

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

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

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

$- 2 t^{2} - t + 1 = 0$ $-2t^2-t+1=0$
$t_{1} = - 1$ $t_1=-1$
$t_{2} = \frac{1}{2}$ $t_2=1/2$

but $sin (x) = t$ $sin(x)=t$

And in your interval:

$sin (x) = - 1$ $sin(x)=-1$ and $x = - \frac{π}{2}$ $x=-pi/2$
$sin (x) = \frac{1}{2}$ $sin(x)=1/2$ and $x = \frac{π}{6} and x = 5 \frac{π}{6}$ $x=pi/6 and x=5pi/6$

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

Hope it helps.

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

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