How do you solve $(2sin^2)2x=1$ in the interval 0 to 2pi?

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Answer 1 · 2016-02-21T07:43:59

$x=22.5^@, 67.5^@,112.5^@,157.5^@$ OR

$x=pi/8, (3pi)/8,(5pi)/8, (7pi)/8$

I beleive the problem is to solve for

$2 sin^2 (2x)=1$

divide both sides of the equation by 2

$(2 sin^2 (2x))/2=1/2$

$(cancel2 sin^2 (2x))/cancel2=1/2$

$sin^2 (2x)=1/2$

$sin 2x=+-1/sqrt2$

$2x=sin^-1 (-1/sqrt2)=225^@, 315^@$

$x_1=225/2=112.5^@$

$x_2=315/2=157.5^@$

Also

$2x=sin^-1 (+1/sqrt2)=45^@, 135^@$

$x_3=45/2=22.5^@$

$x_4=135/2=67.5^@$

have a nice day !

How do you solve (2sin^2)2x=1 (2sin^2)2x=1 in the interval 0 to 2pi?