How do you solve $2 {sin}^{2} (2 x) = 1$ $2sin^2(2x)=1$ ?

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How do you solve $2 {sin}^{2} (2 x) = 1$ $2sin^2(2x)=1$ ?

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Answer 1 · 2016-09-12T23:21:22

${sin}^{2} (2 x) = \frac{1}{2}$ $sin^2(2x) = 1/2$

$sin (2 x) = \pm \frac{1}{\sqrt{2}}$ $sin(2x) = +-1/sqrt(2)$

$2 x = \pm {sin}^{- 1} (\frac{1}{\sqrt{2}})$ $2x = +-sin^-1(1/sqrt(2))$

$2x = 135˚, 45˚, 225˚, 315˚$

$x = 67.5˚, 22.5˚, 112.5˚ and 157.5˚$

Hopefully this helps!

Answer 2 · 2016-09-13T04:09:01

$pi/8. (3pi)/8, (5pi)/8, (7pi)/8$

Explanation:

$2sin^2 2x = 1$
$sin^2 2x = 1/2$
$sin 2x = +- 1/sqrt2 = +- sqrt2/2$
Trig table and unit circle -->

a. $sin 2x = sqrt2/2$ --> $2x = pi/4, and 2x = (3pi)/4$
When $2x = pi/4$ --> $x = pi/8$
When $2x = (3pi)/4$ --> $x = (3pi)/8$

b. $sin 2x = - sqrt2/2$ --> $2x = - (3pi)/4$ , and $x = - pi/4$
Arc $(-3pi)/4$ is co-terminal to arc $(5pi)/4$
Arc $(-pi/4)$ is co-terminal to arc $(7pi)/4$
When $2x = (5pi)/4$ --> $x = (5pi)/8$
When $2x = (7pi)/4$ --> $x = (7pi)/8$
Answers for $(0, 2pi)$
$pi/8, (3pi)/8, (5pi)/8, (7pi)/8$

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