How do you solve ${sin}^{2} x + 2 cos x = 2$ $sin^2x + 2cosx = 2$ over the interval 0 to 2pi?

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Answer 1 · 2018-03-02T09:05:25

$x = 0, 2 π$ $color(blue)(x=0,2pi$

Identity:

${sin}^{2} x + {cos}^{2} x = 1$ $color(red)bb(sin^2x+cos^2x=1)$

${sin}^{2} x + 2 sin x = 2$ $sin^2x+2sinx=2$

Using identity:

$1 - {cos}^{2} x + 2 cos x = 2$ $1-cos^2x+2cosx=2$

Rearranging, simplifying and equating to zero:

${cos}^{2} x - 2 cos x + 1 = 0$ $cos^2x-2cosx+1=0$

This is just a quadratic in $cos x$ $cosx$

Let $cos x = u$ $cosx=u$

$u^{2} - 2 u + 1$ $u^2-2u+1$

Factor:

$(u - 1) (u - 1) = 0 \Rightarrow u = 1$ $(u-1)(u-1)=0=>u=1$

$u = cos x$ $u=cosx$

$cos x = 1$ $cosx=1$

$x = arccos (cos x) = arccos (1) \Rightarrow x = 0, 2 π$ $x=arccos(cosx)=arccos(1)=>color(blue)(x=0,2pi$

How do you solve sin^2x + 2cosx = 2sin2x+2cosx=2sin^2x + 2cosx = 2 over the interval 0 to 2pi?