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does anyone know how to get this question? I got pi/2 and 3pi/3.
√2 sin² -sin = 0 Solve for values of sin. 0<x<2pi
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How about 0.707
igloo myrtilles fourmis
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Forget the sins for the moment and solve it as if it was a simple quadratic equation.
√2 sin²x -sinx = 0
sinx(√2 sinx -1) = 0
sinx = 0 or (√2 sinx -1) = 0
If (√2 sinx -1) = 0, then √2 sinx = 1 and so sinx = 1/√2.
So now we have sinx = 0 or 1/√2.
Therefore, x = 0, π/4, 3π/4, π, and 2π.
0 and 2π may not be answers, depending on whether the limits of the range are included.
Why did the vector cross the road?
It wanted to be normal.
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Right, because .707 is 1/√2.
igloo myrtilles fourmis
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Also half of the square root of 2 is the same as 1 over the square root of 2.
igloo myrtilles fourmis
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