How do you solve $2 {sin}^{2} x = 2 + cos x$ $2sin^2 x=2+cos x$ between 0 and 2pi?

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How do you solve $2 {sin}^{2} x = 2 + cos x$ $2sin^2 x=2+cos x$ between 0 and 2pi?

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Answer 1 · 2016-03-09T06:19:05

$\frac{π}{2}, \frac{2 π}{3}, \frac{4 π}{3}, and \frac{3 π}{2}$ $pi/2, (2pi)/3, (4pi)/3, and (3pi)/2$

Explanation:

Replace in the equation ${sin}^{2} x$ $sin^2 x$ by $(1 - {cos}^{2} x$ $(1 - cos^2 x$
$2 (1 - {cos}^{2} x) = 2 + cos x$ $2(1 - cos^2 x) = 2 + cos x$
$- 2 {cos}^{2} x = cos x$ $- 2cos^2 x = cos x$
$cos x + 2 {cos}^{2} x = cos x (1 + 2 cos x) = 0$ $cos x + 2cos^2 x = cos x(1 + 2cos x) = 0$
a. cos x = 0 --> $x = \frac{π}{2} and x = \frac{3 π}{2}$ $x = pi/2 and x = (3pi)/2$
b. (1 + 2cos x) = 0 --> $cos x = - \frac{1}{2}$ $cos x = - 1/2$ --> $x = \pm \frac{2 π}{3}$ $x = +- (2pi)/3$
The co-terminal arc of - (2pi)/3 is arc (4pi)/3.
Answer for $(0, 2 π)$ $(0, 2pi)$ :
$\frac{π}{2}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{3 π}{2}$ $pi/2, (2pi)/3, (4pi)/3, (3pi)/2$

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How do you solve $2 {sin}^{2} x = 2 + cos x$ $2sin^2 x=2+cos x$ between 0 and 2pi?

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How do you solve $2 {sin}^{2} x = 2 + cos x$ $2sin^2 x=2+cos x$ between 0 and 2pi?