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 !