Solve: 4cos22x=1, where x is a real number x=(6n−4)(2π),n is an interger ;x=(6n−2)(2π),n is an interger x=(6n−4)(3π)
Posted: Thu Jul 14, 2022 4:01 pm
by answerhappygod
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Solve: 4cos22x=1, where x is a real number x=(6n−4)(2π),n is an interger ;x=(6n−2)(2π),n is an interger x=(6n−4)(3π),n is an interger ;x=(6n−2)(3π),n is an interger x=n(3π),n is an interger x=n(2π),n is an interger None of these