Find all values of x in the interval [0,2pi] that satisfy the equation. 16 sin^2 (x) = 8?
2015-08-31 4:45 am
If somebody can explain how to do this question, and provide a correct answer that would really help me out right now. I am super lost with these kinds of equations.
回答 (2)
2015-08-31 4:52 am
16 sin²(x) = 8
sin²(x) = 1/2
sin(x) = 1/√2 or sin(x) = -1/√2
[x = π/4 or x = π - (π/4)] or [x = π + (π/4) or x = 2π - (π/4)]
x = π/4, 3π/4, 5π/4, 7π/4
sin²(x) = 1/2
sin(x) = 1/√2 or sin(x) = -1/√2
[x = π/4 or x = π - (π/4)] or [x = π + (π/4) or x = 2π - (π/4)]
x = π/4, 3π/4, 5π/4, 7π/4
2015-08-31 4:46 am
sinx = 1/sqrt2 ---->x = ..
sinx = -1/sqrt2 ---->x = ...
sinx = -1/sqrt2 ---->x = ...
收錄日期: 2021-04-18 00:07:03
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