What are all the solutions between 0 and 2π for $2 {cos}^{2} = sin x + 1$ $2cos^2 = sinx+1$ ?

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What are all the solutions between 0 and 2π for $2 {cos}^{2} = sin x + 1$ $2cos^2 = sinx+1$ ?

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Answer 1 · 2016-03-29T16:46:09

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

Explanation:

Assuming the equation was meant to be:
$XXX 2 {cos}^{2} (x) = sin (x) + 1$ $color(white)("XXX")2cos^2(color(red)(x))=sin(x)+1$

Using the relationship:
$XXX {cos}^{2} (x) + {sin}^{2} (x) = 1$ $color(white)("XXX")cos^2(x)+sin^2(x)=1$
$2 {cos}^{2} (x) = sin (x) + 1$ $2 cos^2(x)=sin(x)+1$

$\to 2 (1 - {sin}^{2} (x)) = sin (x) + 1$ $rarr 2(1-sin^2(x))=sin(x)+1$

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

$\to (2 sin (x) - 1) (sin (x) + 1) = 0$ $rarr (2sin(x)-1)(sin(x)+1)=0$

$\to sin (x) = \frac{1}{2} XXX or XXX sin (x) = - 1$ $rarr sin(x)=1/2color(white)("XXX")orcolor(white)("XXX")sin(x)=-1$
$\to x = \frac{π}{6} or \frac{5 π}{6} XXXXXXX \to x = \frac{3 π}{2}$ $rarr x=pi/6 or (5pi)/6color(white)("XXXXXXX")rarr x=(3pi)/2$
$XXXXXXXXXX$ $color(white)("XXXXXXXXXX")$ (for $x \in [0, 2 π]$ $x in[0,2pi]$ )

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What are all the solutions between 0 and 2π for $2 {cos}^{2} = sin x + 1$ $2cos^2 = sinx+1$ ?

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Explanation:

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What are all the solutions between 0 and 2π for $2 {cos}^{2} = sin x + 1$ $2cos^2 = sinx+1$ ?