Question:
If cos2x = 0
then how
2x= π/2, 3π/2, 5π/2, 7π/2 ?
Answer:
Given
cos2x = 0
=> cos2x = cos(π/2)
=> 2x = 2*n*π ± (π/2) {(if cos x = cos α => x = 2*n*π ± α), by trigonmetry Equation}
Put n = 0, 1 ,2 ,3....., we get
2x = ± (π/2), 2π ± (π/2), 4π ± (π/2), .........
=> 2x = π/2, 3π/2, 5π/2, 7π/2, 9π/2, ............. (Taken positive and negative signs alternatively)
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