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## #1 2006-06-24 10:00:11

fusilli_jerry89
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### Trig

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

## #2 2006-09-01 19:31:10

John E. Franklin
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### Re: Trig

igloo myrtilles fourmis

## #3 2006-09-01 21:12:49

mathsyperson
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### Re: Trig

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.

## #4 2006-09-01 23:58:49

John E. Franklin
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### Re: Trig

Right, because .707 is 1/√2.

igloo myrtilles fourmis

## #5 2006-09-02 00:01:19

John E. Franklin
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### Re: Trig

Also half of the square root of 2 is the same as 1 over the square root of 2.

igloo myrtilles fourmis