How do you solve $2 - {cos}^{2} (x) = 4 {sin}^{2} (\frac{1}{2} x)$ $2-cos^2(x)=4sin^2(1/2x)$ in the interval [0, 2pi]?

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How do you solve $2 - {cos}^{2} (x) = 4 {sin}^{2} (\frac{1}{2} x)$ $2-cos^2(x)=4sin^2(1/2x)$ in the interval [0, 2pi]?

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Answer 1 · 2016-06-02T03:44:03

$\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4} and \frac{7 π}{4}$ $pi/4, (3pi)/4, (5pi)/4 and (7pi)/4$

Explanation:

Replace in the equation ${cos}^{2} x$ $cos^2 x$ by $(1 - {sin}^{2} x)$ $( 1 - sin^2 x)$ and put the equation in standard form:
$f (x) = 2 - 1 + {sin}^{2} x - 4 {sin}^{2} (\frac{x}{2}) = 0$ $f(x) = 2 - 1 + sin^2 x - 4sin^2 (x/2) = 0$
Replace ${sin}^{2} x$ $sin^2 x$ by $(4 {sin}^{2} (\frac{x}{2}) . {cos}^{2} (\frac{x}{2}))$ $(4sin^2 (x/2).cos^2 (x/2))$
$f (x) = 1 + 4 {sin}^{2} (\frac{x}{2}) . {cos}^{2} (\frac{x}{2}) - 4 {sin}^{2} (\frac{x}{2}) = 0$ $f(x) = 1 + 4sin^2 (x/2).cos^2 (x/2) - 4sin^2 (x/2) = 0$
$f (x) = 1 + 4 {sin}^{2} (\frac{x}{2}) ({cos}^{2} (\frac{x}{2}) - 1) = 0$ $f(x) = 1 + 4sin^2 (x/2)(cos^2 (x/2) - 1) = 0$
Since $({cos}^{2} (\frac{x}{2}) - 1) = - {sin}^{2} (\frac{x}{2})$ $(cos^2 (x/2) - 1) = - sin^2 (x/2)$ , therefor:
$f (x) = 1 - 4 {sin}^{4} (\frac{x}{2}) = 0$ $f(x) = 1 - 4sin^4 (x/2) = 0$
$4 {sin}^{4} (\frac{x}{2}) = 1$ $4sin^4 (x/2) = 1$
${sin}^{4} (\frac{x}{2}) = \frac{1}{4}$ $sin^4 (x/2) = 1/4$ -->
${sin}^{2} (\frac{x}{2}) = \frac{1}{2}$ $sin^2 (x/2) = 1/2$ --> $sin x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$ $sin x = +- 1/sqrt2 = +- sqrt2/2$
Trig table and unit circle -->
a. $sin x = \frac{\sqrt{2}}{2}$ $sin x = sqrt2/2$ . There are 2 solution arcs:
$x = \frac{π}{4}$ $x = pi/4$ and $x = \frac{3 π}{4}$ $x = (3pi)/4$
b. $sin x = - \frac{\sqrt{2}}{2}$ $sin x = -sqrt2/2$ . Two solution arcs:
$x = \frac{5 π}{4}$ $x = (5pi)/4$ and $x = \frac{7 π}{4}$ $x = (7pi)/4$

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How do you solve $2 - {cos}^{2} (x) = 4 {sin}^{2} (\frac{1}{2} x)$ $2-cos^2(x)=4sin^2(1/2x)$ in the interval [0, 2pi]?

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