How to solve for x when restricted to interval [0,2pi] tan(x/2)=(2-sqrt(2))/2sinx?

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How to solve for x when restricted to interval [0,2pi] tan(x/2)=(2-sqrt(2))/2sinx?

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Answer 1 · 2018-04-03T04:52:01

$x = \frac{π}{4} and x = \frac{7 π}{4}$ $x=pi/4 andx=(7pi)/4$

Explanation:

$tan (\frac{x}{2}) = \frac{2 - \sqrt{2}}{2 sin x}$ $tan (x/2) = (2 - sqrt2)/(2sin x)$
$\frac{sin (\frac{x}{2})}{cos (\frac{x}{2})} = \frac{2 - \sqrt{2}}{4 sin (\frac{x}{2}) . cos (\frac{x}{2})}$ $(sin (x/2))/(cos (x/2)) = (2 - sqrt2)/(4sin (x/2).cos (x/2))$
Simplify by $cos (\frac{x}{2})$ $cos (x/2)$ and cross-multiply -->
$4 {sin}^{2} (\frac{x}{2}) = 2 - \sqrt{2}$ $4sin^2 (x/2) = 2 - sqrt2$
${sin}^{2} (\frac{x}{2}) = \frac{2 - \sqrt{2}}{4}$ $sin^2 (x/2) = (2 - sqrt2)/4$
$sin (\frac{x}{2}) = \pm \frac{\sqrt{2 - \sqrt{2}}}{2} = \pm 0.38$ $sin (x/2) = +- (sqrt(2 - sqrt2))/2 = +- 0.38$
Calculator and unit circle give 4 solutions-->
a. $sin (\frac{x}{2}) = (\frac{\sqrt{2 - \sqrt{2}}}{2}) = 0.38$ $sin (x/2) = (sqrt(2 - sqrt2)/2) = 0.38$ -->
$\frac{x}{2} = \frac{π}{8} \Rightarrow x = \frac{π}{4} \in [0, 2 π]$ $x/2 = pi/8=>x=pi/4in[0,2pi]$ , and
$x = 2 π - \frac{π}{4} \Rightarrow x = \frac{7 π}{4} \in [0, 2 π]$ $x = 2pi-pi/4 =>x = (7pi)/4in[0,2pi]$
b. $sin (\frac{x}{2}) = - 0.38$ $sin( x/2) = - 0.38$
$x = - \frac{π}{4} \notin [0, 2 π]$ $x = - pi/4!in[0,2pi]$ and
$x = - \frac{7 π}{4} \notin [0, 2 π]$ $x = - (7pi)/4!in [0,2pi]$
Hence,
$x = \frac{π}{4} and x = \frac{7 π}{4}$ $x=pi/4 andx=(7pi)/4$

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How to solve for x when restricted to interval [0,2pi] tan(x/2)=(2-sqrt(2))/2sinx?

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