Asked by Hayli
find the solutions of the equation that are in the interval [0,2pi)
2-cos^2x=4sin^2(1/2x)
2-cos^2x=4sin^2(1/2x)
Answers
Answered by
Reiny
2 - (1 - 2sin^2 (x/2) = 4sin^2 (x/2)
1 = 2 sin^2 (x/2)
sin^2 (x/2) = 1/2
sin (x/2) = 1/√2
x/2 = π/4 or x/2 = 3π/4
then x = π/2 or x = 3π/2
1 = 2 sin^2 (x/2)
sin^2 (x/2) = 1/2
sin (x/2) = 1/√2
x/2 = π/4 or x/2 = 3π/4
then x = π/2 or x = 3π/2
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