Rei A.

asked • 11/15/21

trigonometry questionn

find the solution set of sin2(4x)+5cos2(2X)-2=0 given the interval [0,2pi]


Use only trigonometric identities.



Mark M.

Should sin^2 be of 4x and not 2x?
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11/15/21

Dayv O.

As written in the problem, the function graphs out nicely and has eight zeroes between x=0 and x=pi. Given the calculations simplify so nicely the problem is most likely written perfectly.
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11/15/21

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Dayv O. answered • 11/15/21

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sin2(4x)+5cos2(2X)-2=0

5cos2(2X)=(5/2)(1+cos(4x))

sin2(4x)=1-cos2(4x)

now have quadradic in cos(4x)

cos2(4x)-(5/2)cos(4x)-3/2=0

cos(4x)=3,- or cos(4x)= -(1/2)

only second answer is valid.

x1=(π/6)+πk/2 ,,, k=0,1,2,3 for values in range

x2=(π/3)+πk/2 ,,,,k=0,1,2,3 for values in range.


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