How do you find all solutions of the equation in the interval (0, 2pi) $1/2 = 1 - cos^2 x$ ?

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How do you find all solutions of the equation in the interval (0, 2pi) $1/2 = 1 - cos^2 x$ ?

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Answer 1 · 2016-03-01T05:53:06

$pi/4, (3pi)/4, (5pi)/4, (7pi)/4$

Explanation:

$1/2 = 1 - cos^2 x$ . Replace $(1 - cos^2 x)$ by $sin^2 x.$
$1/2 = sin^2 x$
$sin x = +- sqrt2/2$
a. $sin x = sqrt2/2$ --> $x = pi/4$ and $x = (3pi)/4$
b. $sin x = - sqrt2/2$ --> $x = - pi/4$ and $x = (-3pi)/4$

Note: $(-pi/4)$ --> $(7pi)/4$ (co-terminal) and $- (3pi)/4$ --> $(5pi)/4$ (co-terminal)

Answers for $(0, 2pi)$ :
$pi/4, (3pi)/4, (5pi)/4, (7pi)/4$

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How do you find all solutions of the equation in the interval (0, 2pi) $1/2 = 1 - cos^2 x$ ?

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How do you find all solutions of the equation in the interval (0, 2pi) 1/2 = 1 - cos^2 x1/2 = 1 - cos^2 x?

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How do you find all solutions of the equation in the interval (0, 2pi) $1/2 = 1 - cos^2 x$ ?